1.
The potential difference across the 10 W resistor is 7.1 V.
7.1 V
10 W
12 W
e = 15.0 V
a)
25 W
r
What is the power dissipated by the 25 W resistor?
(4 marks)
V
R
7.1
=
10
I =
I = 0.71 A
¨
1
2
mark
1
1
1
=
+
R | | 12 25
R | | = 8.1 W
¨ 1 mark
V| | = IR | |
= 0.71(8.1)
V| | = 5.76 V
V2
R
5.76 2
=
25
0306phk
¨ 1 mark
1
2
P25 =
¨
P25 = 1.3 W
¨ 1 mark
mark
-1-
July 17, 2003
2.
a)
For the circuit below, what is the terminal voltage of the battery?
(4 marks)
e= 12.0 V
10 W
10 W
r = 4.0 W
R parallel = 5.0 W
Rtotal = 9.0 W
I total =
V 12.0
=
= 1.33 A
R 9.0
Ir drop = 1.33( 4.0) = 5.3 V
Vterminal = 12.0 - 5.3 = 6.7 V
b)
¨ 4 marks
If resistor R is added in parallel to the circuit as shown, what is the effect on the
terminal voltage?
(1 mark)
R
❑
❑
✓
❑
0308phk
increase
no change
decrease
-2-
September 4, 2003
c)
Using principles of physics, explain your choice for b).
(4 marks)
Additional R in parallel results in an overall lower R, thus an increase in current.
(2 marks) As a consequence, a greater voltage drop Ir occurs across the internal
resistance resulting in a smaller terminal voltage. (2 marks)
0308phk
-3-
September 4, 2003
3.
The current through the 50. 0 Ω resistor in the circuit below is 0.14 A.
R1 = 10. 0 Ω
ε
R2 = 33.0 Ω
R3 = 50. 0 Ω
I = 0.14 A
R4 = 68. 0 Ω
r = 0. 80 Ω
a)
Determine the emf of the battery.
(5 marks)
V|| = I ⋅ R
R|| =
= I3 ⋅ R3
= 15. 4 Ω
= 0.14 ⋅ 50. 0
= 7. 0 V
∴ I2 =
=
I4 =
=
← 1 mark
1
2
mark
←
1
2
mark
= 10. 0 + 15. 4 + 0.80
= 26. 2 Ω
7. 0
33. 0
∴ ε = I ⋅ RT
←
1
2
mark
= 0. 45 ⋅ 26. 2
V||
R4
= 11. 9 V
7. 0
68. 0
= 0.10 A
←
RT = R1 + R|| + r
V||
R2
= 0. 21 A
1
1
1
1
+
+
R2 R3 R4
= 12 V
←
1
2
← 1 mark
mark
∴ I|| = I2 + I3 + I4
= 0. 21 + 0.14 + 0.10
= 0. 45 A
0106phk
← 1 mark
-4-
July 25, 2001
4.
In the circuit below, resistor R1 dissipates 0. 40 W . Resistors R2 and R3 are identical.
P1 = 0. 40 W
R1 = 10. 0 Ω
12.0 V
R2
R3
R4 = 15. 0 Ω
What is the resistance of R2 ?
(7 marks)
Key:
P = I2R
P1 = I 2 R1
1
P 2
I =  1
 R1 
1
0. 40  2
= 
10 
= 0. 20 A
V1 = IR
= 0. 2 (10 )
= 2V
V 4 = IR
= 0. 2 (15 )
= 3V
V3 = V 4 = 12 − V1 − V 4
= 7V
I2 = I3
V3 = I3 R3
7 = 0.1 R2
R2 = 70 Ω
0101phk






 ← 2 marks








 ← 1 mark




 ← 1 mark



 ← 1 mark

← 1 mark


 ← 1 mark


-5-
February 23, 2001
5.
A 12 V battery from a car is used to operate a 65 W headlight.
a)
How much energy does the headlight use in 1.5 hours?
E = P×t
←
1
2
(2 marks)
mark
= 65 × 1. 5 × 3 600 ← 1 mark
= 3. 5 × 10 5 J
b)
1
2
mark
What total charge passes through the headlight during this time?
Q=
=
∆E
V
←
3. 5 × 10 5 J
12 V
← 2 marks
= 29 000 C
c)
←
←
1
2
1
2
Q = It
mark
OR
mark
(3 marks)
←
1
2
mark
= ( 5. 42 A )( 5 400 s )
← 2 marks
= 29 000 C
←
1
2
mark
What is the total number of electrons that pass through the headlight during this
time period?
N=
=
Q
e
← 1 mark
29 000
1. 6 × 10 −19 C
← 1 mark
(2 marks)
= 1.8 × 10 23 electrons
0201phk
-6-
March 4, 2002
6.
A 12 V battery transfers 33 C of charge to an external circuit in 7.5 s.
a)
What current flows through the circuit?
I=
=
a
t
←
1
2
mark
33 C
7.5 s
←
1
2
mark
(2 marks)
= 4.4 A ← 1 mark
b)
What is the resistance of the circuit?
R=
=
V
I
←
1
2
mark
12
4.4
←
1
2
mark
(2 marks)
= 2.7 Ω ← 1 mark
c)
What is the power output of the battery?
P =V ⋅I
←
1
2
mark
= 12 4.4
←
1
2
mark
= 53 W
← 1 mark
( )
d)
(2 marks)
The external circuit is most likely to consist of
✔
❐
❐
❐
a bulb.
a kettle.
a calculator.
(Check one response.)
0108phk
(1 mark)
-7-
September 6, 2001
7.
What is the potential difference across the 6. 0 Ω resistor in the circuit shown?
(7 marks)
15. 0 Ω
7. 0 Ω
V = 8.0 V
5.0 Ω
6. 0 Ω
9. 0 Ω
Rp1 = 15. 0 Ω + 6. 0 Ω +9. 0 Ω
= 30. 0 Ω
1 mark
1
1
1
=
+
Rp 7. 0 30. 0
Rp = 5. 68 Ω
1 mark
RT = 5. 0 + 5. 68
= 10. 68 Ω
IT =
VT
8. 0
=
= 0. 75 A
RT 10. 68
1 mark
1 mark
Vp = VT - V5
= 8. 0 V - 0. 75 × 5. 0
= 4. 25 V
Ip =
Vp 4. 25
=
= 0.142 A
Rp 30. 0
1 mark
1 mark
V6 = I p R
= 0.142 × 6. 0
= 0.85 V
0201phk
1 mark
-8-
March 4, 2002
8.
The circuit shown consists of an 8. 00 V battery and two light bulbs. Each light bulb dissipates
5. 0 W. Assume that the light bulbs have a constant resistance. Switch S is open.
Lamp A
ε = 8.00 V
S
r
Lamp B
I = 1.50 A
a)
If a current of 1. 50 A flows in the circuit, what is the internal resistance r of the battery?
(4 marks)
Resistance Solution:
P = I2R
∴ Rbulb =
=
Voltage Solution:
P = IV


5 = 1. 5 V
 ← 1 mark

Vbulb = 3.3 V 
P
I2
5. 0
(1. 50 )2
Vterminal = 3.3 × 2
= 2. 22 Ω ← 1 mark
RT =
Vterminal = 6. 7
Vterminal =
ε − Ir

 ← 1 mark
6. 7 = 8 − 1. 5r 
ε
I
r = 0.89 Ω ← 1 mark
8. 00
=
1. 50
= 12 W
← 1 mark
Pbulbs = 2 ( 5 ) = 10 ← 1 mark
Pr = 12 − 10
← 1 mark
Pr = 2 W
r=
∴ r = RT − 2 ⋅ ( Rbulb )
2
1. 52
= 0.89 Ω
= 5.33 − 2 ( 2. 22 ) ← 1 mark
001phk
= 1. 5( 8 )
P = I2R
= 5.33 Ω ← 1 mark
= 0.89 Ω

 ← 1 mark

Power Solution:
PT = IV
← 1 mark
← 1 mark
-9-
February 24, 2000
b)
The switch S is now closed.
Lamp A
ε = 8.00 V
S
Lamp B
r
Lamp A will now be
i)
4
p
p
p
(1 mark)
brighter.
the same brightness as before.
dimmer.
(Check one response.)
The battery’s terminal voltage will now be
ii)
p
p
4
p
(1 mark)
greater than before.
the same as before.
less than before.
(Check one response.)
c)
Using principles of physics, explain your answers to b).
(3 marks)
Total circuit resistance decreases when the switch is closed. Therefore, the circuit current
increases. ← 1 mark
Since P = I 2 R, the power dissipated by Lamp A increases and it will therefore be
brighter. ← 1 mark
Since the circuit current has increased, the voltage drop across the internal resistance
increases and the terminal voltage drops. ← 1 mark
001phk
- 10 -
February 24, 2000
9.
The terminal voltage of the battery is 5.8 V.
8.0 Ω
S
11 Ω
ε
a)
16 Ω
r = 1.0 Ω
What is the emf of this battery?
(6 marks)
1
1 1
= +
R|| 11 16
R|| = 6.5 Ω
← 1 mark
external RT = 6.5 Ω + 8.0 Ω
external RT = 14.5 Ω
It =
Vt
Rt
5.8
=
14.5
← 1 mark
I t = 0.40 A
← 1 mark
V = ε − Ir
← 1 mark
5.8 = ε − 0.40(1.0)
ε = 6.2 V
b)
← 1 mark
←
1
2
mark
←
1
2
mark
What is the effect on the emf of the battery when switch S is opened?
(1 mark)
no effect
0301phk
-9-
February 25, 2003
10.
How much energy does the 6. 0 Ω resistor dissipate in 15 seconds in the circuit shown?
(7 marks)
15. 0 Ω
V = 12. 0 V
9. 0 Ω
6.0 Ω
5.0 Ω
1
1
1
=
+
Rp 9. 0 6. 0
Rp = 3. 6
← 1 mark
RT = R15 + Rp + R5
= 15. 0 + 3. 6 + 5. 0
= 23. 6 Ω
IT =
← 1 mark
VT 12. 0
=
RT 23. 6
= 0. 508 A
← 1 mark
V p = VT − V15 − V5
= 12. 0 − 0. 51 × 15. 0 − 0. 51 × 5. 0
= 1.83 V
I6 =
← 2 marks
V p 1.83
=
R6
6. 0
= 0.305 A
← 1 mark
E = VIt
= 1.83 × 0.305 × 15
= 8. 4 J
0208phk
← 1 mark
-9-
August 29, 2002
11.
Each of the two cells shown is connected to an external 6.00 Ω resistor.
Cell B
Cell A
ε = 1.5 V
r = 0. 50 Ω
ε = 1.5 V
6.00 Ω
r = 0. 25 Ω
6. 00 Ω
With supporting calculations, state which cell delivers the greater power to the 6.00 Ω resistor.
(7 marks)
Cell A:
I=
I=
Cell B:
ε
6. 00 + r
← 1 mark
I=
1. 5
6. 50
= 0. 23
I=
← 1 mark
6. 00 + r
← 1 mark
1. 5
6. 25
= 0. 24
PL = I 2 R
← 1 mark
PL = I 2 R
= 0. 232 × 6. 00
= 0.32 W
ε
= 0. 24 2 × 6. 00
← 1 mark
= 0.35 W
← 1 mark
Therefore, cell B delivers more power. ← 1 mark
Note: Sig figs were ignored, since answer is not numerical. Also, units were ignored for the
same reason.
0206phk
-8-
July 23, 2002