Simple Circuits - AP Physics C Electricity and Magnetism

In schematic diagrams of electric circuits, connecting wires are typically assumed to be ideal. What is the key property of an ideal wire?
Answer: It has zero resistance, so there is no potential difference between any two points on the same wire. 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.
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?
Answer: When open, there is a potential difference across it; when closed, the potential difference across it is ideally zero. 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.
When the switch $S$ is closed, how does the behavior of the circuit change compared to when the switch is open?
Answer: The resistor $R_2$ is short-circuited, and the total current flowing from the battery increases. 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.
Which resistor in this circuit is a component of more than one fundamental electrical loop?
Answer: $R_1$ 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.
Which of the following best defines an electrical loop within a circuit?
Answer: Any closed path within a circuit through which electric charge can flow and return to its starting point. 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.
If the filament in bulb A breaks, creating an open circuit within that bulb, what happens to bulb B?
Answer: Bulb B immediately goes out because the entire circuit loop is now open. 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.
What is the most immediate consequence of connecting the wire in parallel with the lightbulb?
Answer: The current from the battery will increase significantly, and the lightbulb will go out. 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.
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?
Answer: $\frac{\rho L}{\pi ab}$ 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 $a$ at $x=0$ to $b$ at $x=L$ , we have $r(x) = a + \frac{(b-a)x}{L}$ . The cross-sectional area at position $x$ is $A(x) = \pi r(x)^2$ . 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.
From the perspective of circuit analysis, why is this action considered dangerous for a real battery?
Answer: A real battery has internal resistance; a short circuit causes a very large current to flow, generating significant thermal energy. 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.
Which of the following statements about the potential differences is correct?
Answer: The sum of the potential differences across the resistors is equal to the battery's emf, such that $V_1 + V_2 = V$ . 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 $V$ .
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?
Answer: The current becomes zero, and the potential difference across the switch's contacts becomes equal to the battery's emf. 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.
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?
Answer: $R_L = r; \quad P_{max} = \frac{\mathcal{E}^2}{4r}$ 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.
If the filament in bulb A breaks, creating an open circuit in that branch, what happens to bulb B?
Answer: Bulb B remains at the same brightness because its branch of the circuit is unaffected. 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.
A simple circuit consists of a resistor connected to the terminals of a battery. Which statement accurately describes the flow of charge?
Answer: Conventional current flows from the positive terminal to the negative terminal, while electrons flow from the negative to the positive terminal. 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.
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?
Answer: The open switch creates a break in the conducting path, preventing a complete loop for charge to continuously flow. 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.
A student is asked to build a circuit where a single switch can be used to turn two lightbulbs on or off simultaneously. Which circuit configuration described by a schematic would achieve this?
Answer: A circuit with the two bulbs and the switch all connected in series with the battery. Placing a switch in series with the battery and the bulbs means the switch is part of the main path from the power source. When the switch is open, the entire circuit is broken, and no current flows. When it is closed, the circuit is completed, and current flows to both bulbs. This allows it to control both bulbs simultaneously.
Which of the following statements about the currents is correct?
Answer: The current $I_{bat}$ splits at the junction, such that $I_{bat} = I_1 + I_2$ . 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 ( $I_1 + I_2$ ).
Which of the following descriptions accurately defines an electric circuit?
Answer: A set of electrical components connected by conducting wires to form one or more closed loops for current to flow. 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.
Which of the following correctly pairs a circuit component with its primary function in a simple circuit?
Answer: Resistor: To control the amount of current for a given potential difference and dissipate electrical energy. 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.
In the context of a simple direct current (DC) circuit, what is the primary function of an ideal battery?
Answer: To maintain a constant potential difference between its terminals. 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.

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