Physics 2 Quiz: Resistors In Series And Parallel

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Resistors In Series And ParallelQuestion 1 of 7

An ideal battery of EMF $\mathcal{E}$ is connected to an infinite ladder network of resistors. Each 'rung' of the ladder consists of a resistor $R_s$ in series along the top rail, and a resistor $R_p$ connected as a shunt (in parallel, from top rail to bottom rail). The equivalent resistance of this infinite network as seen from the input terminals is $R_{eq}$ . If $R_s = R_p = R$ , which expression correctly gives $R_{eq}$ ?

R_{eq} = R

, because by symmetry an infinite ladder is self-similar, and this is the only value consistent with

R_{eq} = R_s + \frac{R_p \cdot R_{eq}}{R_p + R_{eq}}

when

R_s = R_p = R

R_{eq} = \frac{R}{2}

, because the series and shunt resistors average out to give an equivalent resistance equal to half of

R

, consistent with the ladder's alternating series-parallel structure.

R_{eq} = \frac{(1+\sqrt{5})}{2}R

, because the self-similar condition

R_{eq} = R + \frac{R \cdot R_{eq}}{R + R_{eq}}

yields a quadratic whose positive root is the golden ratio times

R

R_{eq} = 2R

, because the self-similar equation

R_{eq} = R + \frac{R \cdot R_{eq}}{R + R_{eq}}

can be approximated by treating the shunt resistor as dominant, giving

R_{eq} \approx R + R = 2R

to first order.

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Physics 2 Quiz

Physics 2 Quiz: Resistors In Series And Parallel

Practice Resistors In Series And Parallel in Physics 2 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 Resistors In Series And Parallel, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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

$R_{eq} = R$ , because by symmetry an infinite ladder is self-similar, and this is the only value consistent with $R_{eq} = R_s + \frac{R_p \cdot R_{eq}}{R_p + R_{eq}}$ when $R_s = R_p = R$ .
$R_{eq} = \frac{R}{2}$ , because the series and shunt resistors average out to give an equivalent resistance equal to half of $R$ , consistent with the ladder's alternating series-parallel structure.
$R_{eq} = \frac{(1+\sqrt{5})}{2}R$ , because the self-similar condition $R_{eq} = R + \frac{R \cdot R_{eq}}{R + R_{eq}}$ yields a quadratic whose positive root is the golden ratio times $R$ . (correct answer)
$R_{eq} = 2R$ , because the self-similar equation $R_{eq} = R + \frac{R \cdot R_{eq}}{R + R_{eq}}$ can be approximated by treating the shunt resistor as dominant, giving $R_{eq} \approx R + R = 2R$ to first order.

Explanation: When you encounter an infinite ladder network, the key insight is self-similarity: because the network is infinite, removing one stage leaves a network identical to the original. This means the resistance looking into the network from any stage equals

R_{eq}

itself — a powerful trick that converts an infinite problem into a simple equation. For this ladder, the first series resistor

R_s = R

is followed by

R_p = R

in parallel with the rest of the (identical) infinite network. This gives the self-consistency condition:

R_{eq} = R + \frac{R \cdot R_{eq}}{R + R_{eq}}

Multiplying through by

(R + R_{eq})

R_{eq}(R + R_{eq}) = R(R + R_{eq}) + R \cdot R_{eq}

R_{eq}^2 + R \cdot R_{eq} = R^2 + R \cdot R_{eq} + R \cdot R_{eq}

R_{eq}^2 - R \cdot R_{eq} - R^2 = 0

Using the quadratic formula and taking the positive root:

R_{eq} = \frac{(1 + \sqrt{5})}{2}R

This is exactly the golden ratio

\phi \approx 1.618

times

R

, confirming C is correct. Choice A incorrectly claims

R_{eq} = R

. Plugging

R_{eq} = R

into the self-consistency equation gives

R + \frac{R^2}{2R} = \frac{3R}{2} \neq R

— it simply doesn't satisfy the equation. Choice B (

R_{eq} = R/2

) has no algebraic justification; "averaging" series and parallel elements is not a valid circuit principle. Choice D correctly sets up the equation but then abandons the algebra, replacing an exact quadratic with a rough approximation that doesn't satisfy the original equation. Study tip: Whenever you see an infinite periodic network, immediately apply self-similarity — write one equation, solve the resulting quadratic, and always discard the negative root since resistance must be positive.

Question 2

A circuit consists of a battery with EMF $\mathcal{E}$ and internal resistance $r$ , connected to an external network of resistors. The external network has an equivalent resistance $R_{ext}$ . A student claims that the efficiency of the circuit (defined as the fraction of total power generated by the battery that is delivered to the external network) can be increased without limit by simply adding more resistors in parallel to the external network. Is the student correct, and why?

The student is correct: adding resistors in parallel decreases $R_{ext}$ , which increases total current and total power generated, and since power in the external network scales as $I^2 R_{ext}$ , the efficiency increases continuously as more branches are added.
The student is incorrect: adding resistors in parallel decreases $R_{ext}$ , which decreases efficiency $\eta = \frac{R_{ext}}{r + R_{ext}}$ , because the smaller $R_{ext}$ becomes relative to $r$ , the larger the fraction of power lost in $r$ . Efficiency approaches zero as $R_{ext} \to 0$ . (correct answer)
The student is incorrect: adding resistors in parallel leaves efficiency unchanged, because both the numerator and denominator of $\eta = R_{ext}/(r + R_{ext})$ scale proportionally with any change in $R_{ext}$ , keeping their ratio constant.
The student is correct: adding resistors in parallel increases total current, and since the internal resistance $r$ is fixed, more current through $r$ dissipates more power internally but the external power grows faster, yielding a net efficiency gain that asymptotically approaches 100%.

Explanation: Whenever you see a question about circuit efficiency, your anchor should be the efficiency formula itself:

\eta = \frac{R_{ext}}{r + R_{ext}}

. This tells you immediately that efficiency depends entirely on how

R_{ext}

compares to the fixed internal resistance

r

. Adding resistors in parallel decreases

R_{ext}

, since parallel combinations always yield a resistance smaller than the smallest branch. As

R_{ext}

shrinks, the ratio

\frac{R_{ext}}{r + R_{ext}}

also shrinks — in the limit where

R_{ext} \to 0

, efficiency approaches zero, not 100%. Physically, a tiny external resistance means the battery is nearly short-circuited: almost all power is burned inside

r

, and almost none reaches the load. This confirms that B is correct. Choice A contains a subtle but fatal error: yes, total current increases as

R_{ext}

falls, but tracking current alone is misleading. Power delivered externally is

P_{ext} = I^2 R_{ext}

, and as

R_{ext} \to 0

, this product actually decreases — the shrinking resistance overwhelms the rising current. The efficiency calculation exposes this directly. Choice C is wrong because the ratio

\frac{R_{ext}}{r + R_{ext}}

does not stay constant when

R_{ext}

changes — only the numerator changes while

r

in the denominator remains fixed, breaking any proportionality. Choice D makes the same error as A, incorrectly assuming external power grows faster than internal dissipation when current rises. The math shows the opposite trend. Your study tip: always write out

\eta = \frac{R_{ext}}{r + R_{ext}}

before reasoning about efficiency. The formula instantly reveals whether a proposed change helps or hurts.

Question 3

A network consists of five identical resistors, each of resistance $R$ . Two resistors are connected in series to form branch A. Three resistors are connected in series to form branch B. Branches A and B are then connected in parallel between nodes X and Y. A sixth resistor, also of resistance $R$ , is connected in series with this parallel combination to form the complete circuit driven by an ideal battery of voltage $V$ . What is the power dissipated in the single series resistor (the sixth one)?

$P = \frac{25V^2}{121R}$ , found by computing the equivalent resistance of the entire circuit, determining the total current, and applying $P = I^2 R$ to the sixth resistor. (correct answer)
$P = \frac{36V^2}{121R}$ , found by noting that the sixth resistor carries the full circuit current and computing the current as $\frac{6V}{11R}$ , then applying $P = I^2 R$ .
$P = \frac{V^2}{9R}$ , found by treating the sixth resistor as one of six identical resistors that share the total voltage equally, so each carries one-sixth of $V$ .
$P = \frac{6V^2}{25R}$ , found by computing the fraction of total resistance attributable to the sixth resistor relative to an incorrectly computed equivalent resistance of $\frac{5R}{6}$ .

Explanation: When a circuit mixes series and parallel elements, your first move should always be to reduce the network to a single equivalent resistance, then work outward from there. Start by finding the parallel combination of Branch A (

2R

) and Branch B (

3R

). Parallel resistors combine as

\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{3R} = \frac{5}{6R}

, giving

R_{parallel} = \frac{6R}{5}

. The sixth resistor adds in series, so the total circuit resistance is

R_{total} = R + \frac{6R}{5} = \frac{11R}{5}

. The total current from the battery is

I = \frac{V}{R_{total}} = \frac{5V}{11R}

. Since the sixth resistor is in series with everything else, it carries this full current. Its power is

P = I^2 R = \left(\frac{5V}{11R}\right)^2 R = \frac{25V^2}{121R}

. That confirms A is correct. Choice B makes a subtle but costly error: it claims the total current is

\frac{6V}{11R}

, which would only be true if the total resistance were

\frac{11R}{6}

— a sign the student likely averaged resistances incorrectly rather than applying the proper parallel formula. Choice C assumes all six resistors divide voltage equally, which ignores the fact that parallel branches experience the same voltage, not a shared fraction — a fundamental misapplication of how parallel circuits work. Choice D uses an incorrect equivalent resistance of

\frac{5R}{6}

, which is actually just the parallel combination alone, forgetting to add the sixth resistor back in series before computing power. Study tip: Always resolve the full equivalent resistance before finding current. Errors almost always sneak in when students skip this step or confuse where current splits versus where it stays unified.

Question 4

A circuit has a 36 V ideal battery connected to the following configuration: $R_1 = 9\,\Omega$ is in series with the battery. The far terminal of $R_1$ defines node A; the battery's negative terminal defines node B. Between nodes A and B, resistors $R_2 = 12\,\Omega$ and $R_3 = 6\,\Omega$ are connected in series ( $R_2$ from A to intermediate node C, $R_3$ from C to B). Also between nodes A and B, $R_4 = 18\,\Omega$ is connected directly.

What is the voltage at node C with respect to node B (i.e., $V_C - V_B$ )?

$V_C - V_B = 8\,\text{V}$ , found by using the full 36 V EMF in a voltage-divider ratio involving $R_3$ and the total circuit resistance.
$V_C - V_B = 12\,\text{V}$ , found by computing the voltage across the parallel combination and applying the ratio $R_2/(R_2+R_3)$ to find the drop across $R_2$ , then subtracting from $V_{AB}$ .
$V_C - V_B = 18\,\text{V}$ , found by noting that node C is the midpoint of the dominant current path and assigning it half the total EMF.
$V_C - V_B = 6\,\text{V}$ , found by first computing the voltage $V_{AB}$ across the parallel section, then applying a voltage-divider ratio using $R_3/(R_2+R_3)$ within the series branch from A to B. (correct answer)

Explanation: When a circuit mixes series and parallel resistors, your first move should always be to simplify from the outside in — find the equivalent resistance, then work backward to find individual voltages. Here,

R_2

and

R_3

are in series with each other (total

12 + 6 = 18\,\Omega

), and that series combination is in parallel with

R_4 = 18\,\Omega

. Two equal 18 Ω branches in parallel give

R_{parallel} = 9\,\Omega

. This parallel block is then in series with

R_1 = 9\,\Omega

, so total circuit resistance is

18\,\Omega

. Total current from the battery:

I = 36/18 = 2\,\text{A}

. The voltage across the parallel section (

V_{AB}

) is

2 \times 9 = 18\,\text{V}

. Now, node C sits between

R_2

and

R_3

inside the series branch. Applying a voltage divider within that branch:

V_C - V_B = V_{AB} \times \frac{R_3}{R_2 + R_3} = 18 \times \frac{6}{18} = 6\,\text{V}

. That confirms answer D. Answer A misapplies the voltage divider by using the full 36 V EMF rather than

V_{AB}

, ignoring the drop across

R_1

. Answer B correctly finds

V_{AB} = 18\,\text{V}

but then uses the wrong ratio —

R_2/(R_2+R_3)

gives the drop across

R_2

, not the voltage at node C above B. Answer C arbitrarily assigns half the EMF to node C without any circuit analysis. Study tip: Always reduce the circuit to find

V_{AB}

before applying a voltage divider inside a branch — never divide the full battery voltage unless the branch connects directly across the battery with no other series elements.

Question 5

Two resistors, $R_A$ and $R_B$ , are connected in parallel across an ideal battery. A student claims: 'If $R_A$ is increased, the current through $R_B$ will also decrease, because Kirchhoff's current law requires that changes in one branch affect all branches.' Which of the following correctly evaluates the student's claim?

The student's claim is correct: increasing $R_A$ reduces total current, and by KCL the reduced total current is shared between both branches, so $I_{R_B}$ decreases proportionally with the decrease in total current.
The student's claim is incorrect: in a parallel circuit with an ideal battery, the voltage across $R_B$ is fixed at the battery EMF regardless of $R_A$ , so $I_{R_B} = \mathcal{E}/R_B$ remains constant. KCL is satisfied because the battery adjusts its output current, not by redistributing branch currents. (correct answer)
The student's claim is partially correct: $I_{R_B}$ does not decrease, but it does not remain constant either — it increases slightly because the battery must compensate for the reduced current through $R_A$ by delivering more current to $R_B$ to maintain energy conservation.
The student's claim is incorrect: increasing $R_A$ decreases the equivalent parallel resistance, which increases the total current, and this increased total current causes $I_{R_B}$ to increase rather than decrease.

Explanation: Whenever you see a parallel circuit with an ideal battery, your first instinct should be to ask: what is fixed? An ideal battery maintains a constant terminal voltage — no matter what happens to the external circuit. That single insight unlocks this entire question. Because the battery holds voltage

\mathcal{E}

fixed, every branch in a parallel circuit sees that same voltage independently. The current through

R_B

is simply

I_{R_B} = \mathcal{E}/R_B

. Since neither

\mathcal{E}

nor

R_B

changes when you modify

R_A

, the current through

R_B

stays exactly the same. KCL is still satisfied — the total current does change, but the battery absorbs that change by supplying more or less current from its source. The branch currents are not "redistributed" among each other; instead, the battery adjusts its output. B is correct. A is the student's own claim restated as fact. The flaw is treating the battery like a fixed-current source rather than a fixed-voltage source. Total current does decrease when

R_A

increases, but KCL doesn't force that decrease onto

R_B

— the battery's output current simply decreases. C invents a compensation mechanism that doesn't exist. The battery doesn't "send" extra current to

R_B

; it simply maintains voltage. Energy conservation is already satisfied without redistributing branch currents. D contains a factual error: increasing

R_A

raises the equivalent parallel resistance (

1/R_{eq} = 1/R_A + 1/R_B

gets smaller as

R_A

grows), which decreases total current — the opposite of what D claims. Study tip: On parallel-circuit problems, always anchor your thinking to what is held constant — voltage for an ideal battery, current for an ideal current source. That constraint determines everything else.

Question 6

A circuit contains a battery of EMF $\mathcal{E}$ and internal resistance $r$ . Three resistors are connected to the battery: $R_1$ is connected directly across the battery terminals, and $R_2$ and $R_3$ are connected in series with each other, and this series combination is connected in parallel with $R_1$ . All three resistors have the same resistance $R$ .

If the value of $R$ is doubled while $r$ remains constant, which of the following correctly describes the change in the terminal voltage of the battery?

The terminal voltage increases, because the total external resistance increases, reducing the current drawn from the battery and therefore reducing the voltage drop across the internal resistance. (correct answer)
The terminal voltage decreases, because doubling $R$ increases the resistance of each branch, which reduces the total power delivered to the external circuit and lowers the effective load voltage.
The terminal voltage remains unchanged, because the parallel combination of $R_1$ and the series pair $R_2$ - $R_3$ scales proportionally with $R$ , leaving the ratio of external to internal resistance constant.
The terminal voltage increases initially but then saturates at $\mathcal{E}$ , because as $R \to \infty$ the external resistance dominates and the terminal voltage asymptotically approaches the open-circuit EMF.

Explanation: Whenever you see a question about terminal voltage, anchor your thinking to the fundamental relationship

V_{terminal} = \mathcal{E} - Ir

, where

I

is the current drawn from the battery. Terminal voltage rises when current falls, and falls when current rises. First, work out the external resistance.

R_1

is in parallel with the series combination

R_2 + R_3 = 2R

, giving

R_{ext} = \frac{R \cdot 2R}{R + 2R} = \frac{2R}{3}

. When you double every resistor value,

R_{ext} = \frac{2(2R)}{3} = \frac{4R}{3}

, which is exactly double the original. The total circuit resistance becomes

R_{total} = r + R_{ext}

, so the current is

I = \frac{\mathcal{E}}{r + R_{ext}}

. Doubling

R_{ext}

increases

R_{total}

, which decreases

I

, which decreases the internal voltage drop

Ir

, which increases the terminal voltage. Answer A is correct. Answer B is wrong because it conflates reduced power delivery with reduced terminal voltage — lower current actually means less drop across

r

, so terminal voltage goes up, not down. Answer C is tempting but subtly incorrect: while

R_{ext}

does scale proportionally with

R

, the internal resistance

r

does not scale, so the ratio

R_{ext}/r

changes, and terminal voltage is not preserved. Answer D describes a real physical limit (

V_{terminal} \to \mathcal{E}

R \to \infty

), but the question asks about doubling

R

, a single change — there is no "saturation" behavior to invoke here. Your study tip: always ask "what happens to current?" first. Terminal voltage and current are linked through

r

, so current is the bridge between external resistance changes and terminal voltage changes.

Question 7

A student builds a circuit with a 12 V ideal battery and three resistors: $R_1 = 4\,\Omega$ , $R_2 = 6\,\Omega$ , and $R_3 = 12\,\Omega$ . The student connects $R_2$ and $R_3$ in parallel, and then connects this parallel combination in series with $R_1$ .

The student then removes $R_3$ from the circuit entirely (opening that branch). Which of the following best describes the changes to the voltage across $R_1$ and the current through $R_2$ ?

The voltage across $R_1$ increases and the current through $R_2$ decreases, because removing $R_3$ increases the total resistance, reducing total current; even though $R_2$ now carries all the current, the larger drop across $R_1$ means less voltage is available for $R_2$ .
The voltage across $R_1$ decreases and the current through $R_2$ increases, because the parallel combination's resistance increases, so more of the battery voltage is dropped across the parallel section, leaving less for $R_1$ and driving more current through $R_2$ . (correct answer)
The voltage across $R_1$ increases and the current through $R_2$ remains unchanged, because the total current through the circuit is determined solely by $R_1$ in the series portion and is insensitive to changes in the parallel branch.
The voltage across $R_1$ decreases and the current through $R_2$ decreases, because with $R_3$ removed, the equivalent resistance of the parallel section increases sharply, drawing more voltage and leaving $R_1$ with a proportionally smaller share of the total EMF.

Explanation: Whenever you see a series-parallel circuit with a component removed, your first move should be to recalculate the equivalent resistance and track how voltage divides — because changing one branch reshapes the entire circuit. With all three resistors: The parallel combination of

R_2

and

R_3

gives

R_{23} = \frac{6 \times 12}{6+12} = 4\,\Omega

. Total resistance is

4 + 4 = 8\,\Omega

, so total current is

\frac{12}{8} = 1.5\,\text{A}

. Voltage across

R_1

1.5 \times 4 = 6\,\text{V}

. Voltage across the parallel section:

6\,\text{V}

. Current through

R_2

\frac{6}{6} = 1\,\text{A}

. After removing $R_3$ : The parallel section becomes just

R_2 = 6\,\Omega

. Total resistance is now

4 + 6 = 10\,\Omega

, so total current drops to

\frac{12}{10} = 1.2\,\text{A}

. Voltage across

R_1

1.2 \times 4 = 4.8\,\text{V}

— a decrease. Voltage across

R_2

12 - 4.8 = 7.2\,\text{V}

. Current through

R_2

\frac{7.2}{6} = 1.2\,\text{A}

— an increase. This confirms B. A is wrong because it correctly identifies that total current decreases but incorrectly concludes the voltage across

R_1

increases — the numbers show the opposite. C is wrong because

R_1

's voltage absolutely does change when total resistance changes; no element is "insensitive" to circuit-wide shifts. D gets the direction of

R_1

's voltage change backwards and misattributes the cause. Your strategy: always recompute equivalent resistance before and after any change, then use the voltage divider logic. The series resistor's share of EMF shrinks when parallel resistance grows, because total current falls faster than you might expect.