Module NF2FSLT SP025

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Exercise 33

Determine the equivalent resistances of the resistors in figures (a),
(b) and (c).

2.0  6.0 

2.0  8.0  20  10 
2.0  16  6.0  4.0 
2.0  16 
6.0 
Figure (a) 9.0 
18  Figure (b)

Figure (c)

ANS. : 0.80 ; 2.7 ; 8.0 

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Exercise 34 1.0  7.1 
4.5  3.2 

r 12 V 5.8 

The circuit 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?

ANS. : 1.1 A, 0.3 A; 1.4 A; 11.3 V

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Exercise 35

RR

ε

RR

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.

ANS. : U think

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Exercise 36
Refer to Figure (a) and (b), determine the current I for each case.

5A 5A I

2A

4A 2A
I 1A

Figure (a) Figure (b)

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Solution : Case (a) : Negative sign for I = –3A
indicates that the direction
 Iin   Iout of I shown is wrong. The
52I 4 actual direction of current
I  3A
flow is opposite of that
shown in figure (a).

Case (b) : Positive sign for I = +2A
indicates that the actual
 Iin   Iout direction of I is same as
5  I  21
shown in figure. }
I  2A

LO 3.5 Kirchhoff’s Rules Non-face-to face SLT

Exercise 37

8.50  11.5 V, 2 

15.0 V, 4  6.22 

15.1 

Determine the current and its direction in the
circuit.

LO 3.5 Kirchhoff’s Rules Step 1 : Non-face-to face SLT
Draw
Step 2 : direction ξ Solution
Draw direction I at
8.50  11.5 V, 2  I
each R.
Assume the I flow

anticlockwise

I Loop 1 6.22 

15.0 V, 4 

II

15.1  Step 3 :

Choose direction of

By applying the Kirchhoff’s 2nd law, thus loop (CW @ACW)

Step 4 : ε  IR
Write Kirchoff

2nd law. 15.0  11.5  15.1I  6.22I  2I  8.50I  4I
NOTE : Give

direction I  0.74 A (anticlockwise)
opposite give (–)

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Exercise 38

Referring to the circuit in Figure above,
calculate the current I1, I2 and I3.

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Solution a ce

Step 1 :
Draw direction ξ

I1 Loop 1 Loop 2

I3

b d I3 f Step 3 :

I1 Choose
direction of
Step 2 : color 1 branch loop (CW
path with 1 color. R
on the same branch @ACW)
path will have the
Note : Path dbac : current I1 same value of I.
Path cd : current I2
Path cefd : current I3

LO 3.6 Kirchhoff’s Rules

Non-face-to face SLT

Consider Junction c :  Iin   Iout Step 4 :
I1  I2  I3
Write Kirchoff 2nd law.
I1  I 2  I 3  0 (1) NOTE : if direction of I and ξ

Refer Loop 1 ( acdba – clockwise ) opposite to direction of loop
travel, Give (–) sign

   IR

30 10I1 10I 2  20I1 10I1

40I1 10I 2  30 (2)

Refer Loop 2 ( cefdc– clockwise )

   IR

20 10 10I 3  20I 3 10I 3 10I 2

10I 2  40I 3  10 (3)

LO 3.6 Kirchhoff’s Rules

Non-face-to face SLT

Solve 3 equations simultaneously using scientific calculator

CASIO fx 570 MS CASIO fx 570 ES
Select MODE Select MODE
5 : EQN
EQN 1
Unknowns ? 3 2 : anX+bnY+cnZ = dn

I1 I2  I3  0 (1) a1? = 1, b1 ?= –1, c1? = –1, d1?= 0
40I1 10I 2  30 (2) a2? = 40, b2 ?= 10, c2? = 0, d2?= 30
10I 2  40I 3  10 (3) a3? = 0, b3 ?= –10, c3? = 40, d3?= 10

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

X = 1st unknown (I1);
Y = 2nd unknown (I2);
Z = 3rd unknown (I3)

We get :

I1  0.6666A
I 2  0.3333A
I 3  0.3333A

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Exercise 39

3.9  I1 x 6.7 
I2

12 V 1.2  9.0 V

I1 9.8  I I2



I1 y

For the circuit above, 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 the points x and y.

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Solution Step 1 : 3.9  I1 X I2 6.7  Step 2 :
Draw Draw direction I at
direction ξ Loop 1 I Loop 2
each R.
(a) 1.2 
9.0 V
12 V

Step 3 : 9.8  I1 I I2
Choose direction of
loop (CW @ACW)

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

 Iin  Iout

I1  I2  I I1  I2  I  0

By using the Kirchhoff’s 2nd law,

 From Loop 1:
ε  IR

Step 4 : 12  9.8I1  3.9I1 1.2I

Give direction 13.7I1  1.2I  12 (2)
opposite give (–)

LO 3.6 Kirchhoff’s Rules  From Loop 2: ε  IR Non-face-to face SLT

9.0  6.7I2  1.2I

6.7I2  1.2I  9.0 (3)

SOLVE USING MODE EQN CALCULATOR

CASIO fx 570 ES Unknown letak X = 1st unknown (I1);
Select MODE di kiri Y = 2nd unknown (I2);
5 : EQN Z = 3rd unknown (I)
persamaan,
2 : anX+bnY+cnZ = dn We get : I1 = 0.72 A
nombor di I2 = 1.03 A
sebelah kanan. I = 1.75 A

I1  I2  I  0 a b cd
1 −1 0
13.7I1  1.2I  12 11 0
6.7I2  1.2I  9.0 2 13.7 1.2 12

3 0 6.7 1.2 9

I1  0.72 A; I2  1.03 A; I  1.75 A

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

(b) The potential difference across the 6.7  resistor is given by

V  I2R

V  1.036.7

V  6.90 V

(c) The power dissipated from the 1.2  resistor is

P  I2R

P  1.752 1.2

P  3.68 W

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

(d) The potential difference between the points x and y is

 I1 3.9  12 V 9.8  I1  X

xy

(we travel from y to x)

I

Vxy     IR Y

Vxy  [12  (3.9  9.8)(0.72)] Vxy     IR
Vxy  2.136 V
Vxy  0  (1.75)(1.2)
Vxy  2.1V

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Exercise 40

For a circuit in figure,

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

LO 3.6 Kirchhoff’s Rules Non-face-to face SLT

Exercise 41 4.0 

4.0  5.0 V 5.0 V

10 V 4.0 

Determine the current in each resistor in the circuit shown.

ANS. : 3.75 A; 1.25 A; 1.25 A

LO 3.7 Electrical energy & power Non-face-to face SLT

Exercise 41 r

R

In figure, a battery has an emf of 12 V and an internal
resistance of 1.0 . 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  resistor if the

current flows through it for 20 minutes.

LO 3.7 Electrical energy & power Non-face-to face SLT

Solution

The current in the circuit is given by

  II(5R.0r1).0

12.0 
I  2.0 A

(a) The rate of energy transferred to electrical energy (power)
in the battery is

P  I

P  2.012.0

P  24 W

LO 3.7 Electrical energy & power Non-face-to face SLT

Solution

(b) The rate of heat dissipated due to the internal resistance is

P  I 2r

P  2.02 1.0

P  4.0 W

(c) Given t = 20 minutes = 1200 s

W  I 2Rt

W  (2.0)2 (5.0)1200

W  2.4104 J

LO 3.7 Electrical energy & power Non-face-to face SLT

Exercise 43

(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.

(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 A, 2.0 ; 0.50 

LO 3.7 Electrical energy & power Non-face-to face SLT

Solution

(2) (a) Given :  = 6 V, V= 5 V, R = 10Ω

Find : (i) I = ? (ii) r = ?

(i) V  IR
5  I (10)  I  0.5 A

(ii)   V  I r

6  5  (0.5)r  r  2 

(b) Concept : short circuited means R = 0 Ω

  I(R  r)

1.5  3(0  r)

r  0.5 

LO 3.7 Electrical energy & power Non-face-to face SLT

Exercise 44

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.

th
(College Physics,6 edition, Wilson, Buffa & Lou, Q75, p.589)

4
ANS. : 0.15 A; 1.414  10  m; 2.25 W

LO 3.7 Electrical energy & power Non-face-to face SLT

Solution Given : L = 5.0 m ; d = 3.0 mm = 3.0×10–3 m;
R = 100 Ω ; V = 15 V

Find : (a) I = ? ; (b) ρ = ? ; (c) P = ?

(a) V  IR
15  I (100)  I  0.15 A

(b) R  L  L
d2
A

4

100   (5)   1.414104  m
 (3103)2

4

(c) P  IV  0.15(15)  2.25 W

LO 3.8 Potential divider Non-face-to face SLT
V
Exercise 45

12 V R1  3 
R2  2 

For the circuit above, determine the potential difference
across resistance R2.

LO 3.8 Potential divider Non-face-to face SLT

Solution

Potential difference (voltage) across resistance R2 is given by

V2   R2 V
R1  R2

  2 12
32

 4.8 V

LO 3.8 Potential divider Non-face-to face SLT

Exercise 46

R1 8000 

12 V

R2 4000  Vout

For the circuit above,

a) calculate the output voltage.
b) If a voltmeter of resistance 4000  is connected across

the output, determine the reading of the voltmeter.

LO 3.8 Potential divider Non-face-to face SLT

Solution

(a) The output voltage is given by

Vout   R2 V Vout   4000 12
R1  R2  8000  4000 

Vout  4.0 V

(b) The connection between the voltmeter and 4000  resistor is

parallel, thus the equivalent resistance is

1 1  1 Req  2000 
Req 4000 4000

Hence the new output voltage is given by

Vout   2000 12
 8000  2000 

Vout  2.4 V

Therefore the reading of the voltmeter is 2.4 V.

LO 3.9 Potentionmeter Non-face-to face SLT

Exercise 47

Given that length AB is 1m. When the jockey is tapped at 70 cm from
point A, galvanometer reads zero, determine the value of x.

2V

70 cm B

C
A

G Jockey

+ x-

LO 3.9 Potentiometer Non-face-to face SLT

Solution

As G reads zero   x  VAC 2V

Refer : V  IR  I l A 100 cm B
70 cm
A Concept
V l C ratio

We use ratio to obtain the VAC : 2  100
Use V l VAC 70

Tabulation 2 100 cm VAC  1.4 V
method to
VAC 70 cm
help u

solve it.  X  1.4 V

LO 3.9 Potentionmeter Non-face-to face SLT

Exercise 48 4V

Given that length AB is lAC = ? C Use Tabulation
1m. When the jockey is method to help
tapped at C,
galvanometer reads you solve it.
zero. Determine the
balanced length if value A B
of emf x is 1.8 V.
G Jockey V RL

+- VAB = 4 V LAB = 100
cm
 x=1.8V
ξx =VAC = 1.8 LAC = ?
V   1.8VAs G reads zero  AC V
x

To find lAC, we use ratio : V ∝ l 4 V dibina antara jarak 100 cm

4  100 A B
1.8 lAC C
lAC  45cm
? cm untuk 1.8 V

LO 3.9 Potentionmeter Non-face-to face SLT

Exercise 49

Figure above shows a simple potentiometer. Wire PQ has length 100 cm
and resistance 2.4 Ω. The galvanometer G shows no deflection when the

jockey is at X, a distance 60 cm from P. By neglecting the internal
resistance of the 2.0 V battery, calculate the B of battery B.

(1) Ask as G=0, ξB   VAs G reads zero  BPX Non-face-to face SLT
equal to what
value in the Apply ratio: V ∝ R V R L
circuit below?
P 2.4Ω Q 2V 2.4+1.5 100
(2) Because 2V (driver = 3.9 cm
cell) is not connected 1.5Ω VPQ = 60
directly to the wire 1.23 2.4 cm
PQ but the 2V is 2V VPX =?
divided between wire
PQ and resistance R, 2.0  3.9  VPQ  1.23V
thus we need to VPQ 2.4
calculate use the ratio
methods to obtain the Apply ratio: V ∝ l (3) Use the ration to
voltage across wire obtain emf B .
PQ only.
1.23 V dibina antara jarak 100 cm

PQ
X

? V untuk 60 cm

1.23  100 VPX  0.738V
VPX 60

 B  VPX  0.738V

Example 25

Non-face-to face SLT

PQ is a uniform wire of length 1.0 m and resistance 10.0 .

When the jockey is tapped at C, galvanometer reads zero. Determine emf .

4V , 0.5 As G reads zero   VAC
3
Apply ratio: V ∝ R

10Ω 4  (10  0.5  3) VAB  2.96 V
VAB 10
C B Apply ratio: V ∝ l
A
2.96 V dibina antara jarak 100 cm
58.6 cm
AB
G C

+ -

2.96  100 VAC  1.735V ? V untuk 58.6 cm R L
VAC 58.6
V 100 cm
  VAC  1.735V 58.6 cm
4V 10+0.5+3 =13.5
VAB = 2.96 10

VAC =?

138

CHAPTER 4: MAGNETISM

NON FACE TO FACE
SP025

PHYSICS UNIT
KOLEJ MATRIKULASI JOHOR

LO 4.1 Magnetic Field

Exercise 51

Sketch the magnetic field lines pattern around the bar
magnets for following cases.
a. b.

LO 4.1 Magnetic Field

Solution
(a)

(b)

LO 4.1 Magnetic Field

Exercise 52

Which one of the pictures
shows the magnetic field
that produced by a single
straight current-carrying

wire?

Solution

Answer : P

Exercise 53
LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 53

Two long straight wires are placed parallel to each other and
carrying the same current I. Sketch the magnetic field lines
pattern around both wires

a. when the currents are in the same direction.

b. when the currents are in opposite direction.

Solution I I
a. I

I I

View from above

I

Solution I I
b. I
XI
I

View from above

I

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 54

A long straight wire carries a current of 2.5 A. Find the magnitude of the magnetic
field 25 cm from the wire.

Solution B  oI
2 r
r = 25 cm
 4 107 (2.5)
2 (25102 )

B  2106 T

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 55

A long wire (X) carrying a current of 50 A is placed parallel to and 5.0 cm away
from a similar wire (Y) carrying a current of 10 A.
a. Determine the magnitude and direction of the magnetic flux

density at a point midway between the wires:
i. when the current are in the same direction.
ii. when they are in opposite direction.
b. When the currents are in the same direction there is a point somewhere
between X and Y at which the magnetic flux density is zero. How far from X is
this point?

(Given 0 = 4  107 H m1)

Solution IX  50 A;d  5.0 102 m; IY  10 A

a. i. d

rX A rY rX A rY
IX IY

IX IY rX  rY  d  2.5 102 m
2

By using the equation of magnetic field at any point near the

straight wire, then at point A

Magnitude of BX :

μ0 IX    BX
BX  2πrX  4π 107 50  4.0 104 T
2π 2.5 102

Direction : into the page OR upwards

Solution

a. i. Magnitude of BY :

μ0 I Y    BY
BY  2πrY  4π 107 10
2π 2.5102

BY  8.0 105 T

Direction : out of page OR downwards

Therefore the total magnetic flux density at point A is

 BA  4.0 104  8.0 105
BA  BX  BY BA  3.2 104 T
BA  BX  BY

Direction : into the page OR upwards

Solution d 
 BX
a. ii. rX A rY OR BY
A rY X
I X rX IY

IX IY

By using the equation of magnetic field at any point near the straight

wire, then at point A

Magnitude of BX :

   BX
 4π 107 50 BX  4.0 104 T
2π 2.5102
Direction : into the page OR upwards

Solution IX  50 A;d  5.0102 m; IY 10 A

a. ii. Magnitude of BY : BY  8.0 105 T

 4π 107 10
 BY  2π 2.5102

Therefore the resultant magnetic flux density at point A is
BA  BX  BY

BA  BX  BY BA  4.0 104  8.0 105

BA  4.8 104 T

Direction : into the page OR upwards

Solution d OR 
BX
b. rX C rY I X rX C rY

IY

rX  r

I X I Y rY  d  r

Since the resultant magnetic flux density at point C is zero thus


BC  BX  BY

0  BX  BY BX  μ0 I X and BY  μ0 I Y
BX  BY where 2πrX 2πrY