SF025 STUDENT VERSION

Physics Module SP025 Vol.1 Chapter 3

CHAPTER 3 : ELECTRIC CURRENT & DIRECT CURRENT CIRCUITS

CURRICULUM SPECIFICATIONS

3.1 Electrical Conduction

 Describe microscopic model of current.

*Emphasize on the flow of free electrons in a metal. Include concept of drift

velocity.

 Define electric current, I  dQ .
dt

 Use electric current, I  dQ Q = ne.
,
dt

3.2 Ohm’s law and Resistivity

 State and use Ohm’s Law.

 Define and use resistivity,   RA .
l

 Sketch V-I graph (Experiment 2: Ohm’s Law).
 Verify Ohm’s law (Experiment 2: Ohm’s Law).

 Determine effective resistance of resistors in series and parallel by graphing
method (Experiment 2: Ohm’s Law).

3.3 Variation of resistance with temperature

 Explain the effect of temperature on electrical resistance in metals.

 Use = 0[1 + ( − 0)].
* is at temperature 20oC.

3.4 Electromotive force (emf), internal resistance and potential difference

 Define emf,  and internal resistance, r of a battery.
 State factors that influence internal resistance.
 Describe the relationship between emf of a battery and potential difference across

the battery terminals.
 Use terminal voltage, V    Ir.

3.5 Resistors in series and parallel

 Derive and determine effective resistance of resistors in series and parallel.
3.6 Kirchhoff’s Rules

 State and describe Kirchhoff’s Rules. 70
 Use Kirchhoff’s Rules.

*(i) Maximum two closed circuit loops.
(ii) Use scientific calculator to solve the simultaneous equations.

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3.7 Electrical energy and power
 Use power, P  IV , P  I 2 R and P  V 2 . (Known as power loss).
R
 Use electrical energy, W  IVt

3.8 Potential divider

 Explain principle of potential divider.

 Use equation of potential divider, V1  (  R2 R1  ...Rn )V .
R1  R3

3.9 Potentiometer

 Explain principles of potentiometer and its applications.
 Use related equations for potentiometer, 1  l1 .

2 l2
 Determine internal resistance, r of a dry cell by using potentiometer.

(Experiment 3: Potentiometer).

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CHAPTER 3: ELECTRIC CURRENT & DIRECT CURRENT CIRCUITS

The study of DC (direct current) which is the flowing of current or movement of electric charge
carriers (electrons) due to difference in electrical potential.

3.1 Electrical conduction
Electric current, I

Figure 3.1: Shows a simple closed circuit consists of wires, a battery and a light bulb.

From figure above, the direction of electric field or electric current flow from positive to
negative terminal. Direction of electron flows from negative to positive terminal. The electron
accelerates because of the electric force acted on it.

Electric current, I is defined as the total (nett) charge, Q that flowing through the area per unit
time, t.

Mathematically,

IQ average current
t

I  dQ instantaneous current
dt

It is a base and scalar quantities. The S.I. unit of the electric current is the ampere (A).

1 ampere of current is defined as one coulomb of charge passing through the surface area in
one second OR

1 = 1 = 1 −1
1

Note:
If the charge moves in a circuit at the same direction for all time, the current is called direct
current (dc), which is produced by the battery.

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Electrical conduction in metal

In metal the charge carrier is free electrons and a lot of free electrons are available in it. They
move freely and randomly throughout the crystal lattice structure of the metal but frequently
interact with the lattices. When the electric field is applied to the metal, the freely moving
electron experience an electric force and tend to drift with constant average velocity (called
drift velocity) towards a direction opposite to the direction of the field.

Figure 3.2: Shows constant average velocity (called drift velocity) towards a direction opposite
to the direction of the field.

Note:
The magnitude of the drift velocity is much smaller than the random velocities of the free
electron. Then the electric current is flowing in the opposite direction of the electron flows.

Example 3.1

A silver wire carries a current of 3.0 A. Determine
a. the number of electrons per second pass through the wire
b. the amount of charge flows through a cross-sectional area of the wire in 55 s.

Solution

a. By applying the equation of average current, thus
I  Q and =
t

I  Ne
t

3.0  N (1.601019 )
t

N  1.881019 electronss 1
l

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b. Given t  55s , thus the amount of charge flows is given by

=
= (3)(55)
= 165

Example 3.2

Explain how electrical devices can begin operating almost immediately after you switch
on, even though the individual electrons in the wire may take hours to reach the device.

Solution

Each electron in the wire affects its neighbours by exerting a force on them, causing them
to move. When electrons begin to move out of a battery or source their motion sets up a
propagating influence that moves through the wire at nearly the speed of light, causing
electrons everywhere in the wire begin to move.

3.2 Ohm’s law and Resistivity
Ohm’s law

States that the potential difference across a metallic conductor is proportional to the current
flowing through it if its temperature is constant.

Mathematically,

Where =

Then, (3.1)

V  IR

Where R: resistance of a conductor

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Figures 3.2 (a), 3.2 (b), 3.2 (c) and 3.2 (d): Shows the potential difference V against current I
graphs for various materials.

Note:

Some conductors have resistances which depend on the currents flowing through them are
known as Ohmic conductors and are said to obey Ohm’s law. Meanwhile, non-ohmic

conductors are the conductors where their resistance depends only of the temperature.

Resistivity,ρ

is defined as a measurement of a material’s ability to oppose the flow of an electric current.

Mathematically,

  RA (3.2)
l

Where l : length of the material
A : cross-sectional area

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It is a scalar quantity and its unit is ohm meter ( m). It is also known as specific resistance.
Resistivity depends on the type of the material and on the temperature. A good electric
conductor has a very low resistivities and good insulators have a very high resistivities. From
the eq. (2), the resistance of a conductor depends on the length and cross-sectional area.

Material Resistivity,  (  m)

Silver 1.59  108
Copper 1.68  108
Aluminum 2.82  108
Gold 2.44  108
Glass 10101014

Table 3.1: Shows the resistivity for various materials at T = 20 C.

Example 3.3

Two wires P and Q with circular cross section are made of the same metal and have
equal length. If the resistance of wire P is three times greater than that of wire Q,
determine the ratio of their diameters.

Solution

Given; = = ; = =

= 3 and R ρl , = 2
A
4

 pl p  3 ρQlQ
Ap AQ

l  3 ρl
d 2 d 2
p Q

44

dQ  3 OR dP  1
dP dQ 3

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Example 3.4

An aluminium wire has a resistance of 0.10 Ω. If you draw this wire through a die,
making it thinner and twice as long, what will be its new resistance?

Solution

Given; The initial resistance of the aluminium wire with length and cross-sectional area

Ai is equal to Ri   Li and L f  2Li . Since the density does not change, thus
Ai

A f L f  Ai Li

Af  Ai Li
2Li

Therefore, the final resistance of the wire is given by =



Rf   2Li
Ai

2

Rf  4 Li
Ai

Rf  4Ri

Rf  4(0.10)

R f  0.40

3.3 Variation of resistance with temperature
Effect of temperature on resistance

Metal

When the temperature increases, the number of free electrons per unit volume in metal remains
unchanged. Metal atoms in the crystal lattice vibrate with greater amplitude and cause the
number of collisions between the free electrons and metal atoms increase. Hence the resistance
in the metal increases.

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Temperature coefficient of resistivity, 

is defined as a fractional increase in resistivity of a conductor per unit rise in temperature.

   (3.3)
 0 T

Where = change in resistivity

ΔT = temperature change = Tf − Ti
0 = nitial resistivity

The unit of  is C1 OR K 1.

Since = − 0   0 (1 T ) (3.4)
Then,
Where : final resistivity

From the equation (4), the resistivity of a conductors varies approximately linearly with
temperature.

From the definition of resistivity, thus

R

then the equation (4) can be expressed as

R  R0 (1  T ) (3.5)

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Material  (C1)
Silver
3.80  103
Tungsten 4.50  103
Iron 5.00  103
3.90  103
Aluminum 3.90  103
Copper

Table 3.2: Shows the temperature coefficients of resistivity for various materials at T  200 C

Figure 3.3: Shows the resistance R against temperature T graph for metal

Example 3.5

A copper wire has a resistance of 25m at 200 C . When the wire is carrying a current,
heat produced by the current causes the temperature of the wire to increase by 270 C .

a. Calculate the change in the wire’s resistance.
b. If its original current was 10mA and the potential difference across wire remains

constant, what is its final current?
(Given the temperature coefficient of resistivity for copper is 6.80103 0C 1 )

Solution

a. Given; 0 = 25 × 10−3 ; 0 = 200; = 270

By using the equation for temperature variation of resistance, thus

= 0(1 + )

− 0 = 0 and − 0 =

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R  R0T
R  (25103 )(6.80103 )(27)

R  4.59 103 

b. Given; 0 = 10.0 × 10−3
By using the equation for temperature variation of resistance,

R  R0 (1 T ) where RV and R  V
I I0

V  V (1  T )
I I0

1  1 (1  (6.80103 )(27))
I (10.0 103 )

= 8.45 × 10−3

3.4 Electromotive force (emf), potential difference and internal resistance
Emf,  and potential difference, V

Figure 3.4: Shows a circuit consisting of a battery (cell) that is connected by wires to an
external resistor R.

A current I flows from the terminal A to the terminal B. For the current to flow continuously

from terminal A to B, a source of electromotive force (e.m.f.),  is required such as battery to

maintained the potential difference between point A and point B.

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Electromotive force (emf),  is defined as the energy provided by the source (battery/cell) to

each unit charge that flows through the external and internal resistances.
Terminal potential difference (voltage), V is defined as the work done in bringing a unit (test)
charge from the negative to the positive terminals of the battery through the external resistance
only. The unit for both e.m.f. and potential difference are volt (V). When the current I flows
naturally from the battery there is an internal drop in potential difference (voltage) equal to Ir.

Thus, the terminal potential difference (voltage), V is given by

V    Ir (3.6)

  I(R  r) (3.7)

Where  : e.m.f

V :terminal potential difference(voltage)
Ir :internal drop in potential difference
R :total external resistance

r :internal resistance of a cell (battery)

Equation (6) is valid if the battery (cell) supplied the current to the circuit where

V 

For the battery without internal resistance or if no current flows in the circuit (open circuit),
then equation (6) can be written as

V 

Internal resistance of a battery, r

is defined as the resistance of the chemicals inside the battery (cell) between the poles.

r  Vr
I

Where r :internal resistance
Vr : potential difference across internal resistance

I :current in the circuit

The value of internal resistance depends on the type of chemical material in the battery.

Figure 3.5: Show the symbol of emf and internal resistance in the electrical circuit below.

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Example 3.6

A battery has an emf of 9.0 V and an internal resistance of 6.0 . Determine
a. the potential difference across its terminals when it is supplying a current of
0.5 A.
b. the maximum current which the battery could supply.

Solution

a. Given; = 0.50 By applying the expression for emf, thus

= +

9.0 = + (0.50)(6.0)

= 6.0

b. The current is maximum when the total external resistance, R  0 ,
therefore
= ( + )

9.0  Imax (0  6.0)

I max  1.5A

Example 3.7

A car battery has an emf of 12.0 V and an internal resistance of 1.0 . The external
resistor of resistance 5.0 is connected in series with the battery as shown in Figure
below. Determine the reading of the ammeter and voltmeter if both meters are ideal.

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Solution

Given;   12V ; r  1 ; R  5. By applying the equation of e.m.f., the current in the
circuit is

= ( + )

12 = (5 + 1)

= 2

Therefore, the reading of the ammeter is 2.0 A.

The voltmeter measures the potential difference across the terminals of the battery equal
to the potential difference across the total external resistor, thus its reading is

=

= (2)(5)

= 10

Example 3.8

A battery has an emf of 5.5 V and an internal resistance of 2 is connected to a switch
by a wire of resistance 200 . The voltage across the battery is measured by a voltmeter.
What is the voltmeter reading when the switch is

a. Off
b. On

Solution

a. Given; = 5.5 ; = 2.0 Ω; = 200 Ω . When the switch is off, voltmeter
reading is,
= = 5.5

b. When the switch is on,

= ( + )


= +

=

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= ( + )

5.5
= (200 + 2) 200
= 5.45
Voltmeter reading when the switch is on equals to 5.45V.

3.5 Resistors in series and parallel
Resistors in series

Figure 3.6: Show the symbol of resistor in an electrical circuit.

Figure 3.7: Show three resistors are connected in series to the battery.

Characteristics of resistors in series

The same current I flows through each resistor where
I  I1  I2  I3

Assuming that the connecting wires have no resistance, the total potential difference, V is given
by

V  V1  V2  V3 (3.8)

From the definition of resistance, thus

V1  IR1 V2  IR2 V3  IR3 V  IReff

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Substituting for V1, V2 , V3 and V in the eq. (3.8) gives
IReff  IR1  IR2  IR3

Reff  R1  R2  R3 (3.9)

Where Reff : effective (equivalent) resistance

Resistors in parallel

Figures 3.8 (a) and 3.8 (b): Shows three resistors are connected in parallel to the battery.

Characteristics of resistors in parallel

There same potential difference, V across each resistor where
V  V1  V2  V3

The charge is conserved, therefore the total current I in the circuit is given by

I  I1  I2  I3 (3.10)

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(3.11)
From the definition of resistance, thus

I1  V I2  V I3  V I V
R1 R2 R3 Reff

Substituting for I1, I2 , I3 and I in the eq. (3.10)

V V V V
Reff R1 R2 R3

1 111
= 1 + 2 + 3

Example 3.9

For the circuit in figure above, calculate 86
a. the effective resistance of the circuit.
b. the current passes through the 12 Ω resistor.
c. the potential difference across 4 Ω resistor.
d. the power delivered by the battery.

The internal resistance of the battery may be ignored.

Solution

Given; = ; = = ; =
a. The effective resistance of the circuit,

The resistors R1 and R2 are in series, thus R12 is

R12  R1  R2
R12  4 12

12 = 16Ω

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Since R12 and R3 are in parallel, therefore Reff is given by
1  1 1
Reff R12 R3
1  1 1
Reff 16 2
Reff  1.78

b. Since R1 and R2 are in series, thus

I1  I2  0.50A

c. Hence the potential difference across R1 is

V1  I1R1
V1  (0.50)(4.0)
V1  2.0V

d. The power delivered by the battery is
P V2
Reff
P  (8.0)2
1.78
P  36.0W

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Example 3.10

For the circuit in Figure below, calculate the effective resistance between the points A and
B.

Solution

Given; R1  5 ; R2  5 ; R3  10 ; R4  20 : R5  10

R1 and R2 are connected in series, thus R12 is

R12  R1  R2
R12  5  5  10

Since R1234 and R5 are connected in parallel , therefore the effective resistance Reff is given
by

1  1 1
Reff R1234 R5

1 11
Reff 25 10

Reff  7.14

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3.6 Kirchhoff’s first law (junction or current law)

states the sum of the currents entering any junctions in a circuit must equal the sum of the
currents leaving that junction.

 At any junction: Iin  Iout

Kirchhoff’s second law (loop or voltage law)

states in the algebraic sum of the potential differences in any loop, including those
associated with emf’s and those of resistive elements, must equal zero.

In a closed loop,

   IR OR V  0

Problem solving strategy (Kirchhoff’s Laws)

Step 1: Determine direction of current in the loop.
Choose any one junction and label the current at each junction in the circuit given. Then,
apply the Kirchhoff’s first law.

Method to apply:

At junction A, ∑ = ∑
At junction B, 1 + 2 =

∑ = ∑
= 1 + 2

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Step 2: Determine the sign convention for emf, and product of IR.
Choose any two closed loops in the circuit and designate a direction (clockwise OR
anticlockwise) to travel around the loop in applying the Kirchhoff’s second law.

Sign convention for emf, .

Sign convention for product IR.

Method to apply: = + .
= − .

= + Type equation here.

= − .

Step 3: Solving the simultaneous equation using scientific calculator.
Solving the simultaneous equation to determine the unknown currents and unknown variables.

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Example 3.11

For the circuit in figure below, determine :
a. the currents I1, I2 and I
b. the potential difference across the 6.7 resistor,
c. the power dissipated from the 1.2 resistor,
d. the potential difference between point A and B.

Solution

a. the currents I1, I2 and I

At junction A, by using the Kirchhoff’s 1st law, thus

 Iin  Iout (1)

I1  I2  I (2)
(3)
By using the Kirchoff’s 2nd Law,
91
Loop 1 : V  0
Loop 2 :
12  3.9I1 1.2I  9.8I1  0
13.7I1 1.2I  12

V  0

9  6.7I2 1.2I  0
6.7I2 1.2I  9

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By solving the simultaneous equations, we get :

I1  0.72A I2  1.03A I  1.75A

b. the potential difference across the 6.7 resistor,

V  I2R

= (1.03 )(6.7 )
= 6.90

c. the power dissipated from the 1.2 resistor,
= 2
P  (1.75A) 2 (1.2)
= 3.68

d. the potential difference between point A and B.
= −
= 1(3.9 ) − 12
= −9.192

3.7 Electrical energy and power
Electrical energy and power

Consider a circuit consisting of a battery that is connected by wires to an electrical device
(such as a lamp, motor or battery being charged) as shown in figure 3.9.

Figures 3.9: Show the potential different across that electrical device is V. 92
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A current I flows from the terminal A to the terminal B, if it flows for time t, the charge Q
which it carries from B to A is given by

Q  It

Then the work done on this charge Q from B to A (equal to the electrical energy supplied) is

W  QV

W  VIt (3.12)

Power, P

is defined as the energy liberated per unit time in the electrical device. The electrical power P
supplied to the electrical device is given by

Ps W  Vs It
t t

Ps  IVs (3.13)

Where Ps: power supplied
Vs: voltage supplied by source (eg: battery)

When electric current flows through a resistor, hence the potential difference across the resistor
is given by

VR  IR

Then the electrical power across resistor can be written as

PR  I 2 R PR  V 2 (3.14)
R
OR
R

It is a scalar quantity and its unit is watts (W).
If the electrical device is a passive resistor (device which convert all the electrical energy
supplied into heat), the heat dissipated H is given by

H  W  VR It OR W  I 2 Rt (3.15)

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Example 3.12

In Figure above, a battery has an emf of 12 V and an internal resistance of 1.0 Ω.
Determine :

a. the rate of energy transferred to electrical energy in the battery,
b. the rate of heat dissipated in the battery,
c. the amount of heat loss in the 5.0 Ωresistor if the current flows through it for 20

minutes.

Solution

Given; ;   12V ; r  1 ; R  5 .The current in the circuit is given by

= ( + )

12 = (5 + 1 )

I  2A

a. The rate of energy transferred to electrical energy (power) in the battery is :
=
= (2 )(12 )
= 24

b. The rate of heat dissipated due to the internal resistance is
= 2
P  (2.0A)2 (1)

P  4W

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c. Given = (20)(60) = 1200 . The amount of heat loss in the resistor is:
= 2
= (2 )2(5 )(1200 )
= 2.4 × 104

Combination of cells
Cells in series

Figure 3.10: Show two cells are connected in series

Total emf, = 1 + 2 (3.16)

Total internal resistance, = 1 + 2 (3.17)

Note:
If one cell, e.m.f. 2 say, is turned round ‘in opposition’ to the others, then = 1 − 2 but
the total internal resistance remains unaltered.

Cells in parallel

Figure 3.11: Show two equal cells are connected in parallel

Total emf,  , = 1 (3.18)
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Total internal resistance, r,

1 1 1 (3.19)
r r1 r1

Note:

If different cells are connected in parallel, there is no simple formula for the total emf and the
total internal resistance where Kirchoff’s Laws have to be used.

3.8 Potential divider

A potential divider produces an output voltage that is a fraction of the supply voltage V.

Figure 3.12: Show circuit of potential divider consist of two equal cells.

Since the current flowing through each resistor is the same, thus

= and = 1 + 2



=

1+ 2

Therefore, the potential difference (voltage) 1 across is given

1 = 1 V1  ( R1 )V (3.20)
R1  R2 (3.21)

Similarly,

V2  ( R2 )V
R1  R2

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Physics Module SP025 Vol.1 Chapter 3

Figure 3.13: Show Resistance 1 and 2 can be replaced by a uniform homogeneous wire.

The total resistance, in the wire is

RAB  RAC  RCB and R  l
A

RAB  l1  l2
A A

R AB    l2 )
A (l1

Since the current flowing through the wire is the same, thus

I V
R AB

I  V

 (l1  l2 )
A

Therefore, the potential difference (voltage) across the wire with length l1 is given by

V1  IR AC

V1  (  V )( l1 )
A (l1  l2 ) A

V1  ( l1 )V (3.22)
l1  l2 (3.23)

Similarly,

V2  ( l2 )V
l1  l2

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Note:
From Ohm’s law, V  IR  I ( l )
A
Example 3.13
VI

For the circuit in Figure above, calculate:
a. the output voltage
b. If a voltmeter of resistance 4000 is connected across the output, determine the
reading of the voltmeter.

Solution

a. The output voltage is given by

= ( 1 2 2)
+

Vout  ( 4000 )12
8000 4000

Vout  4.0V

b. The connection between the voltmeter and 4000 resistor is parallel, thus the
equivalent resistance is
111
= 4000 + 4000

Reff  2.4V

Therefore, the reading of the voltmeter is 2.4 V.

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3.9 Potentiometer

Figure 3.14: Show a potentiometer circuit

The potentiometer is balanced when the jockey (sliding contact) is at such a position on wire
AB that there is no current through the galvanometer. Thus, Galvanometer reading = 0. When
the potentiometer in balanced, the unknown voltage (potential difference being measured) is
equal to the voltage across AC.

Vx  VAC

Potentiometer can be used to compare the emfs of two cells, measure an unknown emf of a
cell and to measure the internal resistance of a cell.

Compare the emfs of two cells

Figure 3.14: Show a potentiometer circuit a potentiometer set up to compare the emfs of two
cell in which AB is a wire of uniform resistance and J is a sliding contact (jockey)onto the wire.

An accumulator X maintains a steady current I through the wire AB. Initially, a switch S is
connected to the terminal (1) and the jockey moved until the emf 1 exactly balances the
potential difference (p.d.) from the accumulator (galvanometer reading is zero) at point C.
Hence

1 =

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Where VAC  IRAC and RAC  l1
A

Then 2  ( I )l2 (1)
A

After that, the switch S is connected to the terminal (2) and the jockey moved until the emf 2
balances the p.d. from the accumulator at point D. Hence

= and RAD  l2
A

Then 2  ( I )l2 (2)
A

By dividing eq. (1) and eq. (2) then

1 ( I )l1
2 A

I
( A )l2

1  l1 (3.24)
2 l2

Measure an unknown emf of a cell

By using the same circuit shown in Figure 3.14, the value of unknown emf can be determined
if the cell emf (1) is replaced with a standard cell.
A standard cell is one in which provides a constant and accurately known emf. Thus, the emf(2)
can be calculated by using the equation 3.24.

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Measure the internal resistance of a cell

Figure 3.14: Show a potentiometer circuit with an accumulator of emf  maintains a steady

current I through the wire AB.

Initially, a switch S is opened and the jockey J moved until the emf, 1exactly balances the

emf  from the accumulator (galvanometer reading is zero) at point C. Hence

1  VAC

Where VAC  IRAC and RAC  lo
A

Then 1  ( I )lo (1)
A

After the switch S is closed, the current I1 flows through the resistance box R and the jockey J
moved until the galvanometer reading is zero (balanced condition) at point D as shown in
Figure below.

Figure 3.15: Shows a potentiometer circuit of galvanometer reading is zero (balanced
condition) at point D.

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Hence = and R AD  l
Where = A

Then V  ( I )l (2)
A

From the equation of emf,

1 = + 1

r  1 V and I1  V
I1 R

r  (1 V )R (3)
V

By substituting eqs. (1) and (2) into the eq. (3), we get

r  (lo  l
)R
l

r  (lo 1)R (4)
l

The value of internal resistance, r is determined by plotting the graph of 1/l against 1/R .
Rearranging eq. (4) :

1(r)1  1
l lo R lo

Then compare with y  mx  c

Figure 3.16: Show a straight line graph represent equation 4 102
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Physics Module SP025 Vol.1 Chapter 3

Example 3.14

Cells A and B and centre-zero galvanometer G are connected to a uniform wire OS using
jockeys X and Y as shown in figure above. The length of the uniform wire OS is 1.00 m
and its resistance is 12 . When OY is 75.0 cm, the galvanometer does not show any
deflection when OX= 50.0 cm. If Y touches the end S of the wire, OX = 62.5 cm when
the galvanometer is balanced. The emf of the cell B is 1.0 V. Calculate:
a. the potential difference across OY when Y touches S and the galvanometer is

balanced.
b. the internal resistance of the cell A,
c. the emf of cell A.

Solution

a) Given; lOY1  0.75m ; lOX1  0.50m

Since wire OS is uniform thus
= ( )

ROX  (0.50)12  6.0
1.00

ROY  (0.75)12  9.0
1.00

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When G  0 (balance condition), thus

VOX1   B and VOX1  IROX1

I1ROX1   B

1(6.0) = 1.0
1 = 0.17
Therefore the potential difference across OY is given by

VOY1  I1ROY1  (0.17)(9.0)  1.53V

b) Given; lOY 2  1.00m ; lOX 2  0.625m . Since the wire OS is uniform thus

ROX 2  ( lOX 2 )ROS
lOS

ROX 2  ( 0.625)12  7.5
1.00

ROY 2  (1.00)12  12
1.00

When G  0 (balance condition), thus

VOX 2   B  I 2 ROX 2

I 2 ROX 2   B

2(7.5) = 1.0
2 = 0.13
Therefore the potential difference across OY is given by

VOY  I 2 ROY 2

VOY  (0.13)12

VOY  1.56V

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c) The emf of cell A is given by (1)
(2)
= ( + )

For case in the question (a) :

 A  I1(ROY1  r)

= 0.17(9.0 + )

 A  I 2 (ROY 2  r)

= 0.13(12 + )
(1)=(2)0.17(9.0 + ) = 0.13(12 + )

r  0.65

d) The emf of cell A is

= 0.17(9.0 + )
= 0.17(9.0 + 0.65)
= 1.64

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A. OBJECTIVE QUESTIONS

1. Three resistors, each with resistance R1 are in series in a circuit. They are replaced by
one equivalent resistor R. Comparing this resistor to the first resistor of the initial
circuit, which of the following its true?

A. The current through R equal the current through R1
B. The voltage through R equal the current through R1
C. The power given off by R equal to the power given by R1
D. R is less than R1

2. If the length of a wire of resistance R and its cross-sectional area A are doubled,
which of the followings is true about the magnitude of R and resistivity ?

R 
A. Unchanged Increases
B. Unchanged Unchanged
C. Increases Decreases
D. Increases Unchanged

3. Which of the following parameter DOES NOT alter the electrical resistance of a piece
of metal wire?

A. The temperature of the wire
B. The diameter of the wire
C. The type of material of the wire
D. The direction of current flow in the wire

4. The electromotive force is
A. The potential difference between the terminals of a battery in an open circuit.
B. The force accelerates electrons through a wire when a battery is connected.
C. The energy supplied by a battery for electrons moves through internal and external
resistors.
D. The force that accelerates protons through a wire when a battery is connected.

5. Which statement is FALSE ABOUT Kirchhoff’s Rules?
A. The total internal resistances are equal to the total external resistances in a closed
loop.
B. The total current flow into a junction is equal to the total current flow out of the
junction.
C. The total energy supplied by the batteries is equal to the total potential difference
across resistors.
D. The total potential differences in a closed loop is equal to zero.

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B. STRUCTURE QUESTIONS

1. A current of 4 A flow in the circuit for 3 hours. What is the total charges flow during
this time? (Ans: . × )

2. A potential difference 24 V is applied to 15  resistor. Calculate the magnitude of
current flows through the resistor. (Ans: . )

3. A potential difference of 8  is applied across metal wire with uniform
cross - sectional area of 0.08 2and length of 0.5  . Determine the resistivity of the
wire if the electric current flow is 1.2  . (Ans: . × − Ω )

4. Calculate the equivalent resistance and total current flow from battery for each circuit
below. (Ans: 7.4 Ω, 2.03A,6 Ω, 2A, 7.82 Ω, 5.12 A)

i.

ii.
iii.

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5. A 100 W, 200 V filament lamp has operating temperature of 2000oC. The filament
material has resistance temperature coefficient of 0.005 oC-1 at 20 oC. The current taken

by the lamp at the instant of switching with 200 V supply with filament temperature of
40oC will be (Ans: . )

6. A battery has an emf of 9.0 V and an internal resistance of 6.0 . Determine
a) the potential difference across its terminals when it is supplying a current of 0.50

(Ans: . )
b) the maximum current which the battery could supply. (Ans: . )

7. a) Calculate each of the unknown currents I1, I2, and I3 for the circuit in the
FIGURE 1 below. (Ans: I1=6.52A, I2=14.24 A, I3=20.76 A)

b) Calculate the current through the 12 Ω resistor in the FIGURE 2 circuit by using
Kirchoff’s Law. (Ans: I1= -0.13 A, I2= 0.342 A, I3= -0.355 A)

8. FIGURE 3 shows an arrangement of resistance R1 3.0 and R2 2.0 which are

arrange in parallel to a 12.0 driver source. Determine the potential difference across

resistance R2 . (Ans: 4.8 V)

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9. A battery with an emf of 12  is connected to a 545  resistor. How much energy
disippated in the resistor in 65 s? (Ans: . )

10. A potentiometer consists of a wire CE of 100  in length and 4  resistance. A cell

of 1 is 1.5  and 2.0  internal resistance is connected in series with 2.0  external
resistor. When a cell with emf 2 is connected to the potentiometer, the balance length
CD is 45.5  . Determine 2.

ℰ1= 1.5 V, 2 Ω 2Ω 2
Ω
A B

C D

F ℰ

2

G

C. HOTS QUESTIONS

1.

Figure above shows a rod in is made of two materials. Each conductor has a square
cross section and 3.00 mm on a side. The first material has a resistivity of
4.00  10–3  m and is 25.0 cm long, while the second material has a resistivity of 6.00
 10–3  m and is 40.0 cm long. Determine the resistance between the ends of the rod.
(Physics for scientists and engineers, 6th edition,Serway&Jewett, Q24, p.853)
(Ans: 378 )

2. A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire to the
end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. A
potential difference of 5.0 V is maintained between the ends of the 2.0 m composite
wire. Determine

a) the current in the copper and silver wires.
b) the magnitude of the electric field in copper and silver wires.

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Physics Module SP025 Vol.1 Chapter 3

c) the potential difference between the ends of the silver section of wire.
(Given  (silver) is 1.47  108  m and  (copper) is 1.72  108  m)
(University physics, 11th edition, Young & Freedman, Q25.56, p.976)
(Ans: 45 A, 2.76 V m1, 2.33 V m1, 2.79 V)

3. Determine the equivalent resistances of the resistors in Figures below.
(Ans: 0.80 , 2.7 , 8.0 )

4.

1.0  7.1 
4.5  3.2 

r 12 V 5.8 

The circuit in figure above includes a battery with a finite internal resistance, r = 0.50.
a) Determine the current flowing through the 7.1  and 3.2  resistors.
b) How much current flows through the battery?
c) What is the potential difference between the terminals of the battery?

(Physics, 3th edition, James S. Walker, Q 39, p. 728)

(Ans: 1.1 A, 0.3 A, 1.4 A, 11.3 V)

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5.

Four identical resistors are connected to a battery as shown in Figure above. When the
switch is open, the current through the battery is I0.
a) When the switch is closed, will the current through the battery increase, decrease or

stay the same? Explain.
b) Calculate the current that flows through the battery when the switch is closed, Give

your answer in terms of I0.
(Physics, 3th edition, James S. Walker, Q45, p.728) (Ans: U think)

6. ε1

R1 I1 ε2 R2

I2

R3

I

Given 1= 8V, R2= 2 , R3= 3 , R1 = 1  and I = 3 A. Ignore the internal resistance in
each battery. Calculate
a) the currents I1 and I2.
b) the emf, 2.
(Ans: 1.0 A, 4.0 A, 17 V)

4.0 
7.

4.0  5.0 V 5.0 V

10 V 4.0 

Determine the current in each resistor in the circuit shown in Figure above.
(College Physics, 6th edition, Wilson, Buffa & Lou, Q57, p.619)

(Ans: 3.75 A, 1.25 A, 1.25 A)

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8. A wire of unknown composition has a resistance of 35.0  when immersed in the
water at 20.0 C. When the wire is placed in the boiling water, its resistance rises to
47.6 . Calculate the temperature on a hot day when the wire has a resistance of
37.8 .
(Physics, 7th edition, Cutnell & Johnson, Q15, p.639)
(Ans: 37.8 C)

9. a) A battery of emf 6.0 V is connected across a 10  resistor. If the potential
difference across the resistor is 5.0 V, determine
i. the current in the circuit.
ii. the internal resistance of the battery.
(Ans: 50 A, 2.0 )

b) When a 1.5 V dry cell is short-circuited, a current of 3.0 A flows through the
cell. What is the internal resistance of the cell? (Ans: 0.50 )

10. An electric toy of resistance 2.50  is operated by a dry cell of emf 1.50 V and an
internal resistance 0.25 .
a) What is the current does the toy drawn?
b) If the cell delivers a steady current for 6.00 hours, calculate the charge pass
through the toy.
c) Determine the energy was delivered to the toy.
(Ans: 0.55 A, 1.19  104 C, 16.3 kJ)

11. A wire 5.0 m long and 3.0 mm in diameter has a resistance of 100 . A 15 V of
potential difference is applied across the wire. Determine
a) the current in the wire.
b) the resistivity of the wire.
c) the rate at which heat is being produced in the wire.
(College Physics, 6th edition, Wilson, Buffa & Lou, Q75, p.589)
(Ans: 0.15 A, 1.40  104  m, 2.30 W)

12. In Figure below, PQ is a uniform wire of length 1.0 m and resistance 10.0 .

ε1 S1

R1

Q T P
ε2 G

R2
S2

1 is an accumulator of emf 2.0 V and negligible internal resistance. R1 is a 15 
resistor and R2 is a 5.0  resistor when S1 and S2 open, galvanometer G is balanced

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when QT is 62.5 cm. When both S1 and S2 are closed, the balance length is 10.0 cm.
Calculate
a) the emf of cell 2.
b) the internal resistance of cell 2.
c) the balance length QT when S2 is opened and S1 closed.
d) the balance length QT when S1 is opened and S2 closed.
(Ans: 0.50 V, 7.5 , 25.0 cm, 25.0 cm)

13. A potentiometer with slide-wire of length 100 cm and resistance of 5.0 , is connected
to a driver cell of emf 2.0 V and negligible internal resistance. Calculate

a) the length of the potentiometer wire needed to balance a potential difference of
1.5 V.

b) the resistance which must be connected in series with the slide-wire to give a
potential difference of 7.0 mV across the whole wire.

c) the emf  of a dry cell which is balanced by 80 cm of the wire, setup as in part (b).
(Ans: 75.0 cm, 1424 , 5.6 mV)

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