1. Circuits
Practice Questions

2. Review Formula’s
Charging
Discharging

3. Question
A wire made of brass and another wire made of silver have the same length, but the diameter of the brass wire is 4 times the diameter of the silver wire. The resistivity of brass is 5 times greater than the resistivity of .
Silver. If RB denotes the resistance of the brass wire and RS denotes the resistance of the silver wire, which of the following is true?
RB=5/16 RS
RB=4/5 RS
RB=5/4 RS
RB=5/2 RS
RB=16/5 RS
Let ρs denote the resistivity of silver and let As denote the cross-sectional area of the silver wire.

4. Question
For a ohmic conductor, doubling the voltage without changing the resistance will cause the current to?
Decrease by a factor of 4
Decrease by a factor of 2
Remain unchanged
Increase by a factor of 2
Increase by a factor of 4
Therefore doubling the voltage, doubles the current.

5. Question
If a 60 watt light bulb operates at a voltage of 120V, what is the resistance of the bulb? 2Ω
30Ω
240Ω
720Ω
7200Ω

6. Question
A battery whose emf is 40V has an internal resistance of 5Ω. If this battery is connect to a 15Ω resistor R, what will the voltage drop across R be?
10V
30V
40V
50V
70V

7. Question
Three resistors are connected to a 10-V battery as shown below. What is the current through the 2.0 Ω resistor?
4.0Ω
2.0Ω
ε=10V
0.25A
0.50A
1.0A
2.0A
4.0A
Since all resistors are in series, the amount of current that passes through any one of them is the same. So we need to simply the circuit to determine that current.

8. Question Determine the equivalent resistance between points a and b?
0.167Ω
0.25 Ω
0.333 Ω
1.5 Ω
2 Ω
12Ω 3Ω 4Ω 3Ω

9. Question
Three identical light bulbs are connected to a source of emf, as shown in the diagram above. What will happen if the middle bulb burns out?
All the bulbs will go out
The light intensity of the other two bulbs will decrease (but they won’t go out).
The light intensity of the other two bulbs will increase.
The light intensity of the other two bulbs will remain the same.
More current will be drawn from the source emf.
If each bulb has a resistance of R, then each individual bulb will draw ε/R. This will be unchanged if any individual bulb goes out. (less current will be drawn from the battery, but the same amount of current will pass through each bulb)

10. Question
An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected.
a) Is Higher
b) Is Lower
c) Is The Same
d) Don’t know
Bulbs in parallel are like resistors in parallel. Therefore since the total resistance of parallel resistors is lower, and the voltage ins the same, then the current must increase (double).

11. Question
An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected.
a) Is Higher
b) Is Lower
c) Is The Same
d) Don’t know
Since the total resistance goes up by two, the current must go down by two. Therefore lower.

12. Question
An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the power output from the battery (compared to when only one bulb was connected)
a) Is four times higher
b) Is twice as high
c) Is the same
d) Is half as much
e) Is one quarter as much
f) Don’t know
Since the current goes up by two. Therefore the Power goes up by two

13. Question
An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the light from the first bulb (compared to when only one bulb was connected)
a) is four times as bright
b) is twice as bright
c) is the same
d) is half as bright
e) is one quarter as bright
Since the voltage goes down by a factor of two, the power goes down by 4. Therefore dimmer.

14. Question
What is the voltage drop across the 12 ohm resistor in the portion of the circuit shown?
24V
36V
48V
72V
144V 8Ω
Since the top branch has 4Ω of resistance and the bottom branch has 12Ω of resistance, therefore 3 times as much current will flow through the 4Ω resistor than the 12Ω resistor. This breaks down the 12A into 9A up and 3A down. So form V=IR we have V=(3A)(12Ω)=36V 4Ω 2Ω 8Ω
12A
12Ω
We could also have produced a RT and found 3Ω value, therefore a voltage drop of (3Ω)(12A)=36V on both the upper and lower branch

15. Question
What is the current through the 8Ω resistor in the circuit shown?
0.5A
1.0A
1.25A
1.5A
3.0A 2Ω 2Ω 2Ω
Since points a and b are grounded, they are all at the same potential (call it zero). Travelling from b to a across the battery , the potential increases by 24V, therefore it must decrease by 24V across the 8Ω resistor as we reach point a. Therefore I=V/R =(24V)/(8Ω)=3A. 2Ω 2Ω 2Ω 2Ω
24V 2Ω 8Ω a b

16. Question
How much energy is dissipated as heat in 20 seconds by a 100 Ω resistor that carries a current of 0.5A?
50 J
100 J
250 J
500 J
1000J

17. Question What is the time constant for the circuit shown? 0.01 s
200Ω
50Ω
200 uF
50V

18. Question
A 100Ω, 120Ω, and 150Ω resistor are connected to a 9-V battery as in the circuit shown below. Which of the three resistors dissipates the most power?
We can also answer this question faster by noticing that the voltage drops across the 100Ω and then across the parallel resistors. Since the 100Ω has a higher value than the parallel combination, it will have a larger voltage drop than the combination so via P=IV, it dissipates the most power.
The 100Ω resistor
The 120Ω resistor
The 150Ω resistor
Both the 120Ω and the 150Ω
All dissipate the same power
100Ω
120Ω 9V
150Ω
Therefore the 100Ω resistor dissipates the most amount of power.

19. Question
A 1.0 F capacitor is connected to a 12 V power supply until it is fully charged. The capacitor is then disconnected from the power supply, and then is used to power a toy car. The average drag force on the car is 2 N. about how far will the car go?
36m
72m
144m
24m
12m
This energy is the Work used in moving the car a fixed distance.
First we find the energy stored in the capacitor

20. Question
Three capacitors are connected to a 9 V power supply as shown. How much charge is stored by this system of capacitors?
3 uC
30 uC
2.7 uC
27 uC
10 uC
Simplify the circuit to find equivalent capacitance.
C1=2uF
C2=4uF
C3=6uF 9V
To determine the charge , we use Q=CV

21. Question
What is the resistance of an ideal ammeter and an ideal voltmeter?
Measures current
Measures Voltage
An ammeter is placed in series with other circuit components. In order for the ammeter not to itself resist current and change the total current in the circuit, you want it to have zero resistance.
Ideal Ammeter Ideal Voltmeter
zero
infinite 1Ω
infinite
zero 1Ω
A voltmeter is placed in parallel with other circuit components. If it had a low resistance, the current will flow through it instead of the other circuit element. So you want it to have infinite resistance to it won’t affect the circuit element being measured.

22. Question
A light bulb is rated at 100 W in Canada, where the standard wall outlet voltage is 120V. If this bulb was plugged in in england, where standard wall outlet voltage is 240V, which of the following would be true?
The bulb would be ¼ as bright.
The bulb would be ½ as bright.
The bulb’s brightness would the same.
The bulb would be trice as bright.
The bulb would be 4 times as bright
Your first instinct is to say that because brightness depends on power, the bulb is exactly as bright. But that is not correct. The power of the bulb can change.
The resistance of a light bulb is a property of the bulb itself, and so will not change no matter what the bulb is hooked to.

23. Question
A current of 6.4A flows in a segment of copper wire. The number of electrons crossing a cross-sectional area of the wire every second is about?
6.4
4 x 1019
4 x 10-19
6.4 x 1019
6.4 x 10-19
Since each electron carries a charge of 1.6 x C. The we calculate that

24. Question
In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors
What is the resistance of the toaster and blender at their rated voltages?
Determine the current that is achieved in the blender when it is working properly.
Create a circuit that will make both devices work simultaneously.
What power must the battery supply to run your circuit?

25. Question
In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors
What is the resistance of the toaster and blender at their rated voltages?
Since we know the Power requirements , let’s us a power formula:

26. Question
In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors
Determine the current that is achieved in the blender when it is working properly.
This is a job for Ohm’s Law:

27. Question
In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors
Create a circuit that will make both devices work simultaneously.
Required to give a current of 2A for blender
60Ω
blender
Toaster
ε=120V

28. Question
In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors
What power must the battery supply to run your circuit?
From Ohm’s Law, we can determine current in toaster
Since the blender had a current of 2A, this gives the circuit a total current of 12A
Using P=IV, we have:

29. Question Given the following circuit 10Ω a 10Ω 20Ω 40Ω ε=120V 100Ω b
At what rate does the battery deliver energy to the circuit?
Determine the current through the 20 Ω resistor.
i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher?
Determine the energy dissipated by the 100 Ω resistor in 10 s
Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is Ωm, determine its radius.

30. Question Given the following circuit
At what rate does the battery deliver energy to the circuit?
10Ω
20Ω
100Ω
40Ω
ε=120V
Recall: Rate is Power RT
We need to write this as a simple circuit with a RT so that we can determine the current, I by using V=IR.

31. Question Given the following circuit
b) Determine the current through the 20 Ω resistor.
We can determine the voltage drop .
Recall: from Question a) we have a 2A current
10Ω 2A
10Ω
120V-(2A)(10Ω)-(2A)(10Ω)-(2A)(10Ω)
=60V
20Ω
40Ω
The ratio of the resistance of left side to right side is 40:120 or 1:3
ε=120V
100Ω
Then we use V=IR to find each individual current. 2A
10Ω
Therefore ¼ of 2 A goes through the right and ¾ of 2A goes through the left.
Thus (1/4)(2A)=0.5A passes through the 20Ω resistor. OR

32. Question Given the following circuit
c) i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher?
10Ω
20Ω
100Ω
40Ω
ε=120V a b i)
ii) Point a is at a higher potential. Since current flows from a high potential to a low potential.

33. Question Given the following circuit
d) Determine the energy dissipated by the 100 Ω resistor in 10 s
10Ω
20Ω
100Ω
40Ω
ε=120V a b
Recall: Energy equals Power multiplied by Time

34. Question Given the following circuit
e) Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is Ωm, determine its radius.
20Ω
100Ω
10Ω
40Ω
ε=120V a b

35. Question
Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants.
10Ω R C r ε S b a
Determine the current through r at time t=0.
Compute the time required for the charge on the capacitor to reach one-half its final value.
When the capacitor is fully charged, which plate is positive?
Determine the electric potential energy stored in the capacitor when the current r is zero. When the current through r is zero, the switch S is moved to b [t=0].
Determine the current through R as a function of time.
Find the power dissipated in R as a function of time.

36. Question
Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants.
Determine the current through r at time t=0.
10Ω R C r ε S b a

37. Question
Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants.
Compute the time required for the charge on the capacitor to reach one-half its final value.
10Ω R C r ε S b a
Therefore we need

38. Question
Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants.
When the capacitor is fully charged, which plate is positive?
10Ω R C r ε S b a +
Because the top plate is connected to the positive end of the battery, the top is positive.

39. Question
Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants.
Determine the electric potential energy stored in the capacitor when the current r is zero.
10Ω R C r ε S b a
When the current through r is zero, the capacitor is fully charged, with voltage across the plates matching the emf of the battery.

40. Question
Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants.
When the current through r is zero, the switch S is moved to b [t=0].
Determine the current through R as a function of time.
10Ω R C r ε S b a
The current established by the discharging capacitor decreases exponentially.

41. Question
Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants.
When the current through r is zero, the switch S is moved to b [t=0].
Find the power dissipated in R as a function of time.
10Ω R C r ε S b a

42. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
Simplify the above circuit so that it consists of one equivalent resistor and the battery.
What is the total current through this circuit?
Find the voltage across each resistor.
Find the current through each resistor.
The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same.

43. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
REQ
Simplify the above circuit so that it consists of one equivalent resistor and the battery.

44. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
REQ
=342.2Ω
b) What is the total current through this circuit?

45. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
Find the voltage across each resistor.
Find the current through each resistor.
Let’s use a VIR chart V I R
200Ω
300Ω
400Ω
500Ω
Total 9V
0.0263A
342.2Ω
3.16V
0.0158A
3.16V
0.0105A
5.84V
0.0146A
5.84V
0.0117A

46. Question
200Ω
e) The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same.
300Ω
ε= 9V
400Ω
500Ω
By removing a resistor from a parallel set, we actually increase the resistance of the total circuit.
Therefore by Ohm’s law if the voltage remains the same and the resistance increases, the total current must decrease
Now through the 200Ω set, the total resistance remains the same, yet the current decreases, therefore the voltage across each resistor decreases as well as the current .