Resistors in Series & Parallel - Physics 2

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 $R_2$**?
   Answer: 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 $R_2$. 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.
2. 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?
   Answer: 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. 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.
3. 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?
   Answer: 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. 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 ($V_{terminal} \to \mathcal{E}$ as $R \to \infty$), 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.
4. 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?
   Answer: 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 $R_{ext} \to 0$. 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.
5. 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)?
   Answer: $P = \frac{25V^2}{121R}$, found by computing the equivalent resistance of the entire circuit, determining the total current, and applying $P = I^2 R$ to the sixth resistor. 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.
6. 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 $R_{eq}$?
   Answer: $R_{eq} = \frac{(1+\sqrt{5})}{2}R$, 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 $R$. 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.
7. What is the voltage at node C with respect to node B (i.e., $V_C - V_B$)?
   Answer: $V_C - V_B = 6\,\text{V}$, 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. 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.

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