AP Physics C Electricity and Magnetism Quiz: Simple Circuits

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Simple CircuitsQuestion 1 of 20

A circuit is constructed with an ideal battery of emf $V$ and two resistors, $R_1$ and $R_2$ , connected in series. A switch $S$ is connected in parallel with resistor $R_2$ .

When the switch $S$ is closed, how does the behavior of the circuit change compared to when the switch is open?

The resistor

R_2

is short-circuited, and the total current flowing from the battery increases.

The resistor

R_1

is short-circuited, and the total current flowing from the battery decreases.

Both resistors are short-circuited, and the total current flowing from the battery becomes zero.

The total resistance of the circuit increases because a new path is added, and the total current decreases.

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Simple Circuits

Practice Simple Circuits in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Simple Circuits, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A circuit is constructed with an ideal battery of emf $V$ and two resistors, $R_1$ and $R_2$ , connected in series. A switch $S$ is connected in parallel with resistor $R_2$ .

When the switch $S$ is closed, how does the behavior of the circuit change compared to when the switch is open?

The resistor $R_2$ is short-circuited, and the total current flowing from the battery increases. (correct answer)
The resistor $R_1$ is short-circuited, and the total current flowing from the battery decreases.
Both resistors are short-circuited, and the total current flowing from the battery becomes zero.
The total resistance of the circuit increases because a new path is added, and the total current decreases.

Explanation: Closing the switch creates a zero-resistance path in parallel with

R_2

. This is a short circuit, and nearly all current will bypass

R_2

. The total resistance of the circuit decreases from

R_1 + R_2

to just

R_1

. Since the total resistance decreases, the total current from the battery (

I = V/R_{eq}

) increases.

Question 2

For a steady conventional current to exist in a circuit, which of the following conditions must be met?

The circuit must contain at least one closed conducting loop that includes a source of electromotive force. (correct answer)
The circuit must contain at least one switch that is in the open position to regulate the flow.
The net charge within the wires of the circuit must be constantly increasing over time.
The circuit must be connected to a ground point to allow for the continuous dissipation of charge.

Explanation: A steady current requires a continuous, unbroken path for charge to flow. This is known as a closed loop. Additionally, a source of electromotive force (like a battery) is needed to provide the potential difference that drives the current.

Question 3

A student builds a simple circuit with a 9V battery and a 10 $\Omega$ resistor. The student then accidentally connects an ideal wire directly across the terminals of the battery.

From the perspective of circuit analysis, why is this action considered dangerous for a real battery?

A real battery has internal resistance; a short circuit causes a very large current to flow, generating significant thermal energy. (correct answer)
The wire adds significant resistance to the circuit, causing the battery to overheat as it tries to push current through.
Connecting the wire in this way reverses the polarity of the battery, which can damage the internal chemical structure.
The electric field inside the wire becomes zero, which causes a rapid buildup of static charge on the battery terminals.

Explanation: A real battery has a small internal resistance. When its terminals are short-circuited by a low-resistance wire, the total resistance of the circuit is just the internal resistance, which is very small. This results in a very large current (

I = \mathcal{E}/r_{int}

). The power dissipated as heat inside the battery (

P = I^2 r_{int}

) becomes very large, which can cause overheating, leakage, or explosion.

Question 4

A simple circuit consists of an ideal battery connected to a lightbulb. A conducting wire with negligible resistance is then connected in parallel with the lightbulb.

What is the most immediate consequence of connecting the wire in parallel with the lightbulb?

The current from the battery will increase significantly, and the lightbulb will go out. (correct answer)
The current through the lightbulb will double because there are now two paths for the current.
The potential difference across the lightbulb will increase, causing it to become brighter.
The total resistance of the circuit will increase, causing the current from the battery to decrease.

Explanation: The wire with negligible resistance creates a 'short circuit' across the lightbulb. Since current follows the path of least resistance, almost all the current will flow through the wire, bypassing the bulb. This causes the bulb to go out and the total resistance of the circuit to drop to near zero, resulting in a very large current from the battery.

Question 5

A circuit contains an ideal battery, a resistor, and a switch, all connected in series. The switch is initially closed, and a steady current is flowing.

If the switch is opened, what are the immediate effects on the current in the circuit and the potential difference across the switch's contacts?

The current becomes zero, and the potential difference across the switch's contacts becomes equal to the battery's emf. (correct answer)
The current becomes zero, and the potential difference across the switch's contacts remains zero.
The current is halved, and the potential difference across the switch's contacts is half the battery's emf.
The current remains unchanged for a brief moment, and the potential difference across the switch's contacts is zero.

Explanation: Opening the switch breaks the conducting loop, so the steady current immediately ceases and becomes zero. With no current flowing through the resistor, there is no potential drop across it (

\Delta V = IR = 0

). Therefore, the full potential difference of the battery's emf appears across the gap created by the open switch.

Question 6

Two identical lightbulbs, A and B, are connected in series to an ideal battery. The circuit is complete, and both bulbs are glowing.

If the filament in bulb A breaks, creating an open circuit within that bulb, what happens to bulb B?

Bulb B immediately goes out because the entire circuit loop is now open. (correct answer)
Bulb B becomes brighter because it now receives the full potential difference of the battery.
Bulb B remains at the same brightness because its operation is independent of bulb A.
Bulb B becomes dimmer but stays lit because some current can still flow through the circuit.

Explanation: In a series circuit, there is only one path for the current to flow. If the filament in bulb A breaks, it creates a gap in this path, resulting in an open circuit. Current cannot flow anywhere in an open circuit, so bulb B will also go out.

Question 7

Two identical lightbulbs, A and B, are connected in parallel to an ideal battery. The circuit is complete, and both bulbs are glowing.

If the filament in bulb A breaks, creating an open circuit in that branch, what happens to bulb B?

Bulb B remains at the same brightness because its branch of the circuit is unaffected. (correct answer)
Bulb B goes out because the circuit is no longer complete.
Bulb B becomes brighter because it no longer has to share current with bulb A.
Bulb B becomes dimmer because the total resistance of the circuit has increased.

Explanation: In a parallel circuit, each branch is connected directly across the terminals of the battery and provides an independent path for current. The potential difference across bulb B is determined by the battery and does not change when bulb A's branch is opened. Since the potential difference and resistance of bulb B are unchanged, its current and brightness remain the same.

Question 8

A circuit contains a battery and three resistors, $R_1$ , $R_2$ , and $R_3$ . $R_1$ is in series with the parallel combination of $R_2$ and $R_3$ .

Which resistor in this circuit is a component of more than one fundamental electrical loop?

$R_1$ (correct answer)
$R_2$
$R_3$
No resistor is part of more than one fundamental loop.

Explanation: One fundamental loop can be traced from the battery, through

R_1

, through

R_2

, and back to the battery. A second fundamental loop can be traced from the battery, through

R_1

, through

R_3

, and back to the battery. Since

R_1

is on the path for both of these loops, it is a component of more than one loop.

R_2

and

R_3

are each part of only one of these two fundamental loops.

Question 9

A simple circuit contains an ideal battery connected to two resistors, $R_1$ and $R_2$ , in parallel. Let $I_1$ be the current through $R_1$ , $I_2$ be the current through $R_2$ , and $I_{bat}$ be the total current leaving the positive terminal of the battery.

Which of the following statements about the currents is correct?

The current $I_{bat}$ splits at the junction, such that $I_{bat} = I_1 + I_2$ . (correct answer)
The current is the same through both resistors and the battery, such that $I_{bat} = I_1 = I_2$ .
The current from the battery flows entirely through the resistor with the smaller resistance value.
The current $I_{bat}$ is equal to the average of the currents through the two resistors.

Explanation: This is a statement of charge conservation at a junction (Kirchhoff's junction rule). The total current entering the junction where the parallel branches begin (

I_{bat}

) must equal the total current leaving it, which is the sum of the currents in the individual branches (

I_1 + I_2

Question 10

A simple circuit contains an ideal battery with emf $V$ connected to two resistors, $R_1$ and $R_2$ , in series. Let $V_1$ be the potential difference across $R_1$ and $V_2$ be the potential difference across $R_2$ .

Which of the following statements about the potential differences is correct?

The sum of the potential differences across the resistors is equal to the battery's emf, such that $V_1 + V_2 = V$ . (correct answer)
The potential difference is the same across both resistors, regardless of their resistance values, such that $V_1 = V_2$ .
The potential difference across each resistor is equal to the battery's emf, such that $V_1 = V_2 = V$ .
The potential difference across the series combination is the product of the individual potential differences.

Explanation: This is a statement of energy conservation in a circuit loop (Kirchhoff's loop rule). The total potential rise provided by the battery (

V

) must equal the sum of the potential drops across all components in the series loop. Therefore, the potential drops across the resistors,

V_1

and

V_2

, must sum to

V

Question 11

A power supply has a fixed EMF $\mathcal{E}$ and a fixed internal resistance $r$ . It is connected to a variable load resistor $R_L$ . The value of $R_L$ is adjusted to maximize the power dissipated by the load resistor. Which of the following gives the correct value of $R_L$ and the corresponding maximum power $P_{max}$ dissipated in it?

$R_L = r; \quad P_{max} = \frac{\mathcal{E}^2}{4r}$ (correct answer)
$R_L = r; \quad P_{max} = \frac{\mathcal{E}^2}{2r}$
$R_L = 0; \quad P_{max} = \frac{\mathcal{E}^2}{r}$
$R_L = 2r; \quad P_{max} = \frac{2\mathcal{E}^2}{9r}$

Explanation: When you encounter a problem about maximizing power transfer from a source to a load, you're dealing with the maximum power transfer theorem — a fundamental principle in circuit analysis that appears frequently on the AP Physics C E&M exam. To find the optimal load resistance, start by writing the power dissipated in the load resistor. The current through the circuit is

I = \frac{\mathcal{E}}{R_L + r}

, so the power in the load is

P_L = I^2 R_L = \frac{\mathcal{E}^2 R_L}{(R_L + r)^2}

. To maximize this power, take the derivative with respect to

R_L

and set it equal to zero. Using the quotient rule:

\frac{dP_L}{dR_L} = \frac{\mathcal{E}^2[(R_L + r)^2 - R_L \cdot 2(R_L + r)]}{(R_L + r)^4} = 0

. This simplifies to

(R_L + r)^2 = 2R_L(R_L + r)

, which gives us

R_L = r

. Substituting back:

P_{max} = \frac{\mathcal{E}^2 \cdot r}{(r + r)^2} = \frac{\mathcal{E}^2}{4r}

. This confirms choice A is correct. Choice B has the right resistance but wrong power — it's missing the factor of 4 in the denominator. Choice C represents a common misconception: while

R_L = 0

maximizes current, it doesn't maximize power in the load (the load dissipates zero power!). Choice D gives

R_L = 2r

, which would actually reduce the power transfer efficiency. Study tip: Remember that maximum power transfer occurs when the load resistance equals the source resistance. This is different from maximum efficiency, which occurs when the load resistance is much larger than the source resistance.

Question 12

A component is fabricated from a material of uniform resistivity $\rho$ . The component has a length $L$ and a circular cross-section. Its radius varies linearly with position $x$ along its length, from $r=a$ at $x=0$ to $r=b$ at $x=L$ . What is the electrical resistance of this component?

$\frac{\rho L}{\pi (\frac{a+b}{2})^2}$
$\frac{\rho L}{\pi ab}$ (correct answer)
$\frac{\rho L}{\pi \sqrt{ab}}$
$\frac{2 \rho L}{\pi (a^2+b^2)}$

Explanation: When you encounter a resistor with varying cross-sectional area, you can't use the simple formula

R = \rho L/A

because the area changes along the length. Instead, you need to integrate by considering the resistance of infinitesimal segments. First, establish how the radius varies with position. Since it changes linearly from

a

x=0

b

x=L

, we have

r(x) = a + \frac{(b-a)x}{L}

. The cross-sectional area at position

x

A(x) = \pi r(x)^2

. For a thin slice of thickness

dx

at position

x

, the resistance is

dR = \frac{\rho dx}{A(x)}

. To find the total resistance, integrate:

R = \int_0^L \frac{\rho dx}{\pi r(x)^2}

. Substituting the expression for

r(x)

and performing the integration (using substitution

u = a + \frac{(b-a)x}{L}

), you get

R = \frac{\rho L}{\pi ab}

, which is choice B. Choice A uses the average radius squared, which incorrectly assumes you can simply average the areas. Choice C uses the geometric mean of

a

and

b

, which has no physical basis for this geometry. Choice D appears to use some form of averaging the areas, but with an incorrect factor of 2. Study tip: For any resistor with varying cross-section, always set up the integral

R = \int \frac{\rho dx}{A(x)}

. The key insight is recognizing when direct integration is needed versus when simple formulas apply.

Question 13

In the context of a simple direct current (DC) circuit, what is the primary function of an ideal battery?

To maintain a constant potential difference between its terminals. (correct answer)
To supply a constant amount of electric charge to the circuit.
To provide a constant current to any circuit it is connected to.
To create new electrons and protons to flow through the circuit.

Explanation: An ideal battery is a source of electromotive force (emf), which acts to maintain a constant potential difference across its terminals, regardless of the current it supplies. The current supplied depends on the external circuit's resistance. The battery moves existing charges; it does not create them or supply a fixed amount of charge.

Question 14

Which of the following descriptions accurately defines an electric circuit?

A set of electrical components connected by conducting wires to form one or more closed loops for current to flow. (correct answer)
Any collection of electrical components, such as resistors and capacitors, whether they are connected or not.
A single electrical component, such as a resistor or a battery, that can influence the flow of charge.
The path taken by a single proton as it travels from the positive to the negative terminal of a power source.

Explanation: The definition of an electric circuit requires three key elements: electrical components, connections between them (usually conducting wires), and the formation of at least one closed path or loop that allows for the continuous flow of current.

Question 15

Which of the following best defines an electrical loop within a circuit?

Any closed path within a circuit through which electric charge can flow and return to its starting point. (correct answer)
Any straight-line segment of wire connecting two circuit components.
The entire set of components connected to a single battery, excluding the battery itself.
A point in the circuit where three or more wires connect, which is also known as a junction.

Explanation: An electrical loop is, by definition, any closed conducting path in a circuit. A charge carrier that traverses a loop starts and ends at the same point, having experienced a net change in potential of zero. This concept is fundamental to applying Kirchhoff's loop rule.

Question 16

Which of the following correctly pairs a circuit component with its primary function in a simple circuit?

Resistor: To control the amount of current for a given potential difference and dissipate electrical energy. (correct answer)
Switch: To provide a constant source of potential difference to the circuit when closed.
Ideal Wire: To provide a specific, calibrated amount of potential drop between components.
Battery: To store electric charge on its plates and release it when the circuit is closed.

Explanation: The primary functions of a resistor are to impede the flow of charge (thus controlling current, per

I = V/R

) and to convert electrical potential energy into other forms, typically thermal energy. The other options misrepresent the functions of a switch, wire, and battery.

Question 17

In a simple series circuit containing a battery, a resistor, and a switch, what is a key difference between the switch when it is open and when it is closed?

When open, there is a potential difference across it; when closed, the potential difference across it is ideally zero. (correct answer)
When open, it has zero resistance to allow potential build-up; when closed, it has a significant resistance.
When open, current flows through it by jumping the gap; when closed, current is blocked by it.
When open or closed, the potential difference across it is always equal to the battery's emf.

Explanation: When a switch is closed, it acts as an ideal wire with zero resistance, so the potential difference across it is zero. When it is open, it breaks the circuit. No current flows, so there is no potential drop across the resistor, and the full potential difference of the source appears across the open contacts of the switch.

Question 18

A simple circuit consists of a resistor connected to the terminals of a battery. Which statement accurately describes the flow of charge?

Conventional current flows from the positive terminal to the negative terminal, while electrons flow from the negative to the positive terminal. (correct answer)
Both conventional current and electrons flow from the positive terminal to the negative terminal through the external circuit.
Both conventional current and electrons flow from the negative terminal to the positive terminal through the external circuit.
Conventional current describes the flow of protons from the positive to the negative terminal, while electrons flow in the opposite direction.

Explanation: By convention, the direction of current is defined as the direction that positive charge would flow, which is from a higher potential (positive terminal) to a lower potential (negative terminal). In metallic conductors, the actual charge carriers are electrons, which are negatively charged and thus flow from the negative terminal to the positive terminal.

Question 19

In schematic diagrams of electric circuits, connecting wires are typically assumed to be ideal. What is the key property of an ideal wire?

It has zero resistance, so there is no potential difference between any two points on the same wire. (correct answer)
It has infinite resistance, so it prevents any loss of current along its length.
It has a constant, non-zero resistance that is uniform along its length.
It can only carry current in one direction, acting as a one-way valve for charge.

Explanation: An ideal wire is considered a perfect conductor with zero electrical resistance. According to Ohm's law (

\Delta V = IR

), if the resistance

R

is zero, the potential difference

\Delta V

across any segment of the wire must also be zero, regardless of the current

I

flowing through it.

Question 20

A simple series circuit with a battery and a resistor has an open switch. Why is there no steady flow of charge in the circuit?

The open switch creates a break in the conducting path, preventing a complete loop for charge to continuously flow. (correct answer)
The battery stops producing a potential difference when the circuit is open, conserving its energy.
The free electrons in the wire stop moving entirely and have zero velocity at all times.
The resistance of the air gap in the switch is infinite, which reflects all electrons back to the battery's negative terminal.

Explanation: A steady current requires a continuous, closed conducting path. An open switch introduces a gap (typically filled with air, an insulator) that breaks this path. While the battery still maintains a potential difference, there is no complete loop for the charge carriers (electrons) to circulate.