Q.Kirchhoff's junction rule is a reflection of

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The current from the battery in the given circuit diagram is (Figure: a circuit with a 15V battery having internal resistance 0.5Ω. From the battery's positive terminal, a 2Ω resistor connects to node A. From node A, a 7Ω resistor connects to the top-right corner node. A 6Ω resistor connects node A to node B (the middle vertical branch). A 1Ω resistor connects the top-right corner node to the bottom-right corner node (the right vertical branch). From the battery's negative terminal (through its 0.5Ω internal resistance), an 8Ω resistor connects to node B. From node B, a 10Ω resistor connects to the bottom-right corner node, which connects via the 1Ω resistor back to the top-right corner node, closing the circuit.) (A) 1A (B) 2A (C) 1.5A (D) 3A

   ›Reveal solutionSolution

   The 7+1+10 = 18Ω path and the direct 6Ω resistor are two parallel routes between the same two nodes; combine them, add the series resistances, and divide the EMF by the total resistance to get 1 A.

   Concept and Intuition

   The circuit is a single loop from the battery's terminals through a series-parallel network. Once the two alternate routes between node A and node B are correctly identified as being in parallel (same start and end nodes), the whole circuit collapses to one equivalent series resistance, and the current from the battery follows directly from $I = \varepsilon/R_{total}$.

   Step-by-Step Solution

   1. From node A to node B there are two paths: the direct $6\Omega$ resistor, and A→(7Ω)→top-right corner→(1Ω)→bottom-right corner→(10Ω)→B, i.e. $7+1+10=18\Omega$.
   2. These two paths are in parallel between the same nodes A and B: $R_{AB} = \dfrac{6\times18}{6+18} = \dfrac{108}{24} = 4.5\ \Omega$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The potential difference between points A and B of the adjoining figure is: [FIGURE - a bridge-type resistor network: A connects to B via a $5\Omega$ resistor and to C via a $5\Omega$ resistor; B connects to D via a $5\Omega$ resistor; C connects to D via two $5\Omega$ resistors in series; a diagonal branch from A to D contains a 2V cell] (A) $\dfrac{2}{3}\ V$ (B) $\dfrac{8}{9}\ V$ (C) $\dfrac{4}{3}\ V$ (D) 2 V

   ›Reveal solutionSolution

   With the $2\,\text{V}$ cell across $A$–$D$, the resistor branches divide the voltage so the relevant potential difference is $\tfrac{4}{3}\,\text{V}$.

   The $2\,\text{V}$ cell sits in the diagonal $A$–$D$ branch, so $A$ and $D$ are held at a fixed $2\,\text{V}$ apart. Take $V_A=2\,\text{V}$, $V_D=0$.

   Two resistor paths connect $A$ to $D$:

   • $A\!-\!B\!-\!D$: $5\,\Omega+5\,\Omega=10\,\Omega$, carrying $I_1=\dfrac{2}{10}=0.2\,\text{A}$.
   • $A\!-\!C\!-\!D$: $5\,\Omega+(5+5)\,\Omega=15\,\Omega$, carrying $I_2=\dfrac{2}{15}\,\text{A}$.

   The node potentials (relative to $D$) are:

   $V_B=I_1(5)=1\,\text{V},\qquad V_C=I_2(10)=\frac{2}{15}\times 10=\frac{4}{3}\,\text{V}.$ …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Effective resistance between A & B is ______, where R = $3\ \Omega$ [FIGURE: a resistor network — A connects through resistors R, 2R, 3R in series to a node N; N connects directly (by a plain wire) to B; N also connects through resistors R then 2R in series (a vertical branch) down to another node M; A also connects directly (by a plain wire) down to node M; M connects through resistors 2R, 3R, 5R in series to B] (A) $\frac{5}{3}\Omega$ (B) $5\Omega$ (C) $3\Omega$ (D) $\frac{3}{5}\Omega$

   ›Reveal solutionSolution

   The two "bypass" wires (no resistor) short two pairs of junctions together to the terminals, collapsing the network into three resistor chains all directly in parallel between A and B — giving $R_{eq}=5R/3=5\,\Omega$.

   Concept and Intuition

   A plain wire (zero resistance) connecting two nodes forces them to the same potential — effectively merging them into a single node. Spotting these shorts first is the key simplification step; once the nodes are correctly merged, what looks like a complicated bridge-like network often collapses into simple series/parallel combinations.

   Step-by-Step Solution

   1. Label the top-chain end node $N$ (after A's top-path resistors $R,2R,3R$) and the bottom-chain start node $M$ (after the middle vertical branch).
   2. The wire from $N$ straight to $B$ means $N$ and $B$ are the same node: $N\equiv B$.
   3. The wire from $A$ straight down to $M$ means $A$ and $M$ are the same node: $A\equiv M$.
   4. Now re-examine each resistor chain, which all effectively run directly between the (single) node A and the (single) node B:

   • Top chain: $R+2R+3R = 6R$ (between A and $N\equiv B$) → directly between A and B.
   • Middle vertical chain: $R+2R=3R$ (between $N\equiv B$ and $M\equiv A$) → also directly between A and B (just the opposite direction).
   • Bottom chain: $2R+3R+5R=10R$ (between $M\equiv A$ and B) → directly between A and B.

   5. All three chains (6R, 3R, 10R) are therefore in parallel between the same two terminals A and B: …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.When two wires are connected in the two gaps of a meter bridge, the balancing point is obtained at a distance of 48.4 cm from the left end of the bridge wire. If the wire in the left gap is stretched so that its resistance increases by 3.2% and the wire in the right gap is stretched so that its length increases by 10%, then the new balancing length from the left end of the bridge wire is nearly (A) 33.3 cm (B) 66.6 cm (C) 44.4 cm (D) 55.5 cm

   ›Reveal solutionSolution

   Meter-bridge balance ratio equals the resistance ratio; track how each resistance changes (a direct % change vs. a stretch, which changes $R\propto L^2$) and re-solve for the new balance point. Answer: 44.4 cm.

   Concept and Intuition

   In a meter bridge, $\dfrac{R_{left}}{R_{right}} = \dfrac{l}{100-l}$ where $l$ is the balancing length from the left. When a wire is stretched (not just its resistance quoted directly), its volume stays constant, so if its length grows by a factor $(1+x)$, its cross-section shrinks by the same factor, and since $R = \rho L/A = \rho L^2/(\text{Volume})$, resistance scales as $L^2$ — a 10% length increase means resistance grows by $(1.1)^2 = 1.21\times$, not just $10\%$.

   Step-by-Step Solution

   1. Initial ratio: $\dfrac{R_1}{R_2} = \dfrac{48.4}{100-48.4} = \dfrac{48.4}{51.6}$.
   2. Left gap resistance increases by 3.2% (given directly): $R_1' = 1.032\,R_1$.
   3. Right gap wire is stretched so its length grows by 10%; resistance scales as $L^2$: $R_2' = (1.10)^2 R_2 = 1.21\,R_2$. …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.In the circuit shown, the cells A and B have negligible resistance. For $V_A = 12V$, $R_1 = 500\Omega$ and $R = 100\Omega$, the galvanometer (G) shows no deflection. The value of $V_B$ is: [FIGURE] (a circuit where cell $V_A$ connects through resistor $R_1$ to a node; from that node a galvanometer $G$ connects onward to cell $V_B$; a resistor $R$ connects from the same node down to a common bottom wire that joins the negative terminals of $V_A$, $R$, and $V_B$, forming a bridge-like network) (A) 4V (B) 2V (C) 12V (D) 6V

   ›Reveal solutionSolution

   The circuit is a Wheatstone bridge with batteries in place of resistors; zero galvanometer deflection means the voltage at the middle node is the same from both sides. Using voltage division, $V_B$ must be $2\,\text{V}$, so the correct option is (B).

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   Concept & Intuition: Wheatstone Bridge Symmetry

   When a galvanometer shows no deflection, it means no current flows through it. That can only happen if the two points it connects are at the same electric potential. In a standard Wheatstone bridge, we compare voltages across two resistor pairs. Here, the “resistors” are actually batteries and a single resistor, but the same principle applies: the voltage at the top-middle node (call it $P$) must be equal when measured from the left side (through $R_1$ and $V_A$) and from the right side (through $V_B$ directly). The bottom rail is the common reference (ground). So we simply compute $V_P$ from each side and set them equal.

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   Step-by-step solution

   1. Identify the nodes and reference

      Let the bottom wire be $0\,\text{V}$ (ground). The top-left node (positive terminal of $V_A$) is at $+12\,\text{V}$. The top-right node (positive terminal of $V_B$) is at $V_B$ (unknown). The middle node (between $R_1$ and $G$) is $P$.

   2. No deflection ⇒ no current through G

      If $G$ shows zero deflection, the current through it is zero. That means the voltage at $P$ is the same whether you approach from the left or from the right. In other words, the potential difference across $G$ is zero: $V_P(\text{left}) = V_P(\text{right})$.

   3. Find $V_P$ from the left side

      On the left, $V_A = 12\,\text{V}$ is connected to $P$ through $R_1 = 500\,\Omega$. Since no current flows through $G$, the only path from $P$ to ground is through $R = 100\,\Omega$ (down to the bottom rail). So $R_1$ and $R$ form a voltage divider between $12\,\text{V}$ and $0\,\text{V}$:

   $V_P = 12\,\text{V} \times \frac{R}{R_1 + R} = 12 \times \frac{100}{500+100} = 12 \times \frac{100}{600} = 12 \times \frac{1}{6} = 2\,\text{V}.$

   4. Find $V_P$ from the right side …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.In the circuit shown in the figure, the current (I) is 6 A when $R_3$ is infinite and current (I) is 9 A when $R_3$ is short circuited. Then, the values of $R_1$ and $R_2$ are respectively [FIGURE] (a circuit with a $36V$ battery (current I) connected to two vertical resistors $R_1$ (left) and $R_2$ (right) at the outer edges; between them, in the middle, are two parallel horizontal branches, one containing resistor $R_3$ and the other containing resistor $R_2$) (A) $4\Omega, 2\Omega$ (B) $2\Omega, 4\Omega$ (C) $2\Omega, 2\Omega$ (D) $1\Omega, 4\Omega$

   ›Reveal solutionSolution

   Two current readings give $R_1+2R_2=6\,\Omega$ and $R_1+R_2=4\,\Omega$, so $R_1=R_2=2\,\Omega$.

   Concept and Intuition

   The current path is $R_1$ in series with the parallel pair $(R_3 \parallel R_2^{\text{mid}})$ in series with the right-hand $R_2$. When $R_3=\infty$ the parallel pair reduces to the middle $R_2$; when $R_3=0$ the parallel pair is a short and drops out. Each condition gives a simple series equation from $R_{\text{eq}}=V/I$.

   Step-by-Step Solution

   1. With $R_3=\infty$: $R_{\text{eq}} = \dfrac{36}{6}=6\,\Omega = R_1 + R_2^{\text{mid}} + R_2 = R_1 + 2R_2$.
   2. With $R_3=0$: the middle branch is shorted, so $R_{\text{eq}} = \dfrac{36}{9}=4\,\Omega = R_1 + R_2$.
   3. Subtract: $(R_1+2R_2)-(R_1+R_2)=6-4 \Rightarrow R_2 = 2\,\Omega$. …

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7. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.The equivalent resistance across AB of the circuit is [FIGURE] (a resistor network shaped like a folded triangle: an apex point C at the top connects via a 10Ω resistor down-left and a 10Ω resistor down-right to two upper corner points; the left upper corner connects via a 10Ω resistor down the outer left side to a bottom-left corner; the right upper corner connects via a 10Ω resistor down the outer right side to point B; two 20Ω resistors run diagonally, one from the left upper corner area down to point A and the other from the right upper corner area down to point B; points A and B lie next to each other at the bottom) (A) 20Ω (B) 10Ω (C) 80Ω (D) 40Ω

   ›Reveal solutionSolution

   A symmetric resistor network folds into two identical branches from C to A and from C to B, each reducing to 10Ω by series-parallel combination, giving a total of 20Ω between A and B.

   Concept and Intuition

   When a network has left-right mirror symmetry about a central apex, we can analyze one half and double it (for the identical mirror half) rather than solving the whole bridge at once. Here, each side of the apex C offers two parallel routes down to its respective terminal — a direct 20Ω resistor and a 10Ω+10Ω series path through the upper corner node — and these combine identically on both sides.

   Step-by-Step Solution

   1. On the left side: C connects to upper-left node P via 10Ω, and P connects on down to terminal A via 10Ω (through the outer left side), giving a series path C→P→A of $10+10=20\,\Omega$. C also connects directly to A via a 20Ω diagonal resistor. These two 20Ω paths are in parallel: $\left(\dfrac{1}{20}+\dfrac{1}{20}\right)^{-1}=10\,\Omega$.
   2. By the mirror symmetry of the network, the right side gives the same result: the effective resistance from C to B is also $10\,\Omega$. …

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8. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.When a wire is connected in the left gap of a metre bridge, the balancing point is at 40 cm from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap, then the balancing point from the left end of the bridge wire is (A) $\dfrac{300}{11}\ \mathrm{cm}$ (B) $\dfrac{800}{11}\ \mathrm{cm}$ (C) $\dfrac{400}{11}\ \mathrm{cm}$ (D) $\dfrac{700}{11}\ \mathrm{cm}$

   ›Reveal solutionSolution

   Stretching a wire to double its length quadruples its resistance (constant volume, $R\propto L^2$); reapply the meter-bridge balance condition with the new resistance.

   Concept and Intuition

   In a meter bridge, $\dfrac{X}{Y}=\dfrac{l}{100-l}$ where $X$ is the unknown resistance in the left gap, $Y$ the fixed resistance in the right gap, and $l$ the balance length from the left end. When a wire is stretched, its volume $V=AL$ stays constant, but $A$ decreases as $L$ increases. Since $R=\rho L/A = \rho L^2/(AL)=\rho L^2/V$, resistance is proportional to $L^2$: doubling the length quadruples the resistance.

   Step-by-Step Solution

   1. Initial balance: $l=40$ cm, so $\dfrac{X}{Y}=\dfrac{40}{100-40}=\dfrac{40}{60}=\dfrac{2}{3}$.
   2. After stretching to double the length, volume is conserved, so $X' = 4X$ (since $R\propto L^2$ at constant volume).
   3. New ratio: $\dfrac{X'}{Y} = 4\cdot\dfrac{X}{Y} = 4\cdot\dfrac{2}{3} = \dfrac{8}{3}$.
   4. New balance point $l'$: $\dfrac{X'}{Y}=\dfrac{l'}{100-l'} \Rightarrow \dfrac{8}{3}=\dfrac{l'}{100-l'}$. …

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9. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.In the circuit of resistors shown in the figure, the effective resistance between points A and B is [FIGURE] (a triangular network of resistors: the left side of the triangle is a 50 Ω resistor; from the top vertex two 20 Ω resistors run down to two adjacent internal nodes labelled A and B; the right side of the triangle is a 10 Ω resistor connecting to the right vertex; the base of the triangle is a 10 Ω resistor) (A) 10 $\Omega$ (B) $\dfrac{20}{3}\ \Omega$ (C) 65 $\Omega$ (D) 15 $\Omega$

   ›Reveal solutionSolution

   The five resistors form a balanced Wheatstone bridge between A and B: the two $20\ \Omega$ arms and two $10\ \Omega$ arms give $\frac{20}{10}=\frac{20}{10}$, so the $50\ \Omega$ bridge arm carries no current and $R_{AB}=(20+10)\parallel(20+10)=15\ \Omega$ -> option (D).

   Read the network as a bridge with A and B the input terminals and two intermediate nodes P and Q:

   • Arm A-P: $20\ \Omega$, arm P-B: $10\ \Omega$
   • Arm A-Q: $20\ \Omega$, arm Q-B: $10\ \Omega$
   • Bridge arm P-Q: $50\ \Omega$

   Balance condition.

   $\frac{R_{AP}}{R_{PB}}=\frac{20}{10}=2,\qquad \frac{R_{AQ}}{R_{QB}}=\frac{20}{10}=2.$ …

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10. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.If the Wheatstone bridge shown in the figure is balanced, then the value of R is [FIGURE] (a Wheatstone bridge circuit: junction A connects to junction B through a 200 Ω resistor and to junction D through a 100 Ω resistor; junction B connects to junction C through a resistor R and junction D connects to junction C through a 50 Ω resistor; a galvanometer G is connected between B and D; a battery is connected across A and C through the outer loop) (A) 50 Ω (B) 200 Ω (C) 300 Ω (D) 100 Ω

    ›Reveal solutionSolution

    This is a balanced Wheatstone bridge; applying the balance condition to the four arms gives $R=100\ \Omega$.

    Concept and Intuition

    In a balanced Wheatstone bridge, no current flows through the galvanometer, so the two nodes it connects (here B and D) are at the same potential. This means the potential drop from A to B equals the drop from A to D (in the appropriate ratio), and the same for the drops from B to C and D to C. This lets us relate the four resistances without needing to know the battery EMF.

    Step-by-Step Solution

    1. Since the bridge is balanced, the currents through AB and BC are equal (call it $I_1$, since no current diverts through the galvanometer at B), and the currents through AD and DC are equal (call it $I_2$).
    2. Because $V_B=V_D$: the drop $V_A-V_B$ (across the 200 Ω arm, current $I_1$) equals the drop $V_A-V_D$ (across the 100 Ω arm, current $I_2$):

    $I_1(200)=I_2(100)\ \Rightarrow\ \frac{I_1}{I_2}=\frac{1}{2}.$

    3. Likewise $V_B-V_C=V_D-V_C$: the drop across arm BC ($R$, current $I_1$) equals the drop across arm DC (50 Ω, current $I_2$): $I_1 R = I_2(50).$ …

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11. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.In the given circuit, if the current through 4 Ω resistor is zero, then the value of the resistance R is [FIGURE] (a circuit diagram: a 9 V battery at the bottom left is in series with a 1 Ω resistor, with current i flowing upward into a top-left node; from that node a 1 Ω resistor connects rightward, with a current arrow, to a second top node; from the first top-left node a branch goes down through a 4 Ω resistor in series with a 6 V battery to the bottom rail; from the second top node a branch goes down through a 3 V battery to the bottom rail, and further to the right another branch goes down through resistance R to the bottom rail; all the bottom nodes are connected along a common bottom wire) (A) 1 Ω (B) 2 Ω (C) 3 Ω (D) 4 Ω

    ›Reveal solutionSolution

    Zero current in the 4 Ω branch turns the network into one series loop; combining the loop equation with the branch's zero-current voltage condition pins down $R=2\ \Omega$.

    Concept and Intuition

    When a branch of a network carries zero current, it does not affect the currents elsewhere in the circuit (Kirchhoff's current law lets us simply ignore it when tracing current paths) — but the voltage across that branch is still fixed by whatever component sits in it (here, purely by the 6 V battery's EMF, since there's no IR drop across the 4 Ω resistor when I=0). This gives an extra equation relating the potentials of the two nodes that branch connects to, on top of the main loop equation for the current that does flow.

    Step-by-Step Solution

    1. Since the 4 Ω branch carries no current, the only current path is the outer loop: bottom-left node → 1 Ω → top-left node → 1 Ω → top-middle/top-right node → R → bottom-right node → 3 V battery → bottom-middle node → 9 V battery → bottom-left node. A single current $i$ flows around this loop.
    2. Applying Kirchhoff's voltage law around this loop (resistive drops of $1+1=2\,\Omega$ in series with $R$, balanced against the two battery EMFs in the bottom wire) gives one relation between $i$ and $R$.
    3. The zero-current condition on the middle branch means the potential difference between the top-middle node and the bottom-middle node equals exactly the 6 V battery's EMF (no drop across the 4 Ω resistor since $I=0$ there). …

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12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.In a meter bridge, the null point is located at 20 cm from left end of the wire when resistances R and S are connected in the left and right gaps respectively. If the resistance S is shunted with 60 Ω resistance, the null point shifted by 5 cm, then the values of R and S are respectively (A) 24 Ω, 6 Ω (B) 6 Ω, 24 Ω (C) 5 Ω, 20 Ω (D) 20 Ω, 5 Ω

    ›Reveal solutionSolution

    This tests the meter bridge (Wheatstone bridge) balance condition and how shunting one arm shifts the null point; the answer is (C) $R=5\ \Omega,\ S=20\ \Omega$.

    Concept and Intuition

    A meter bridge balances when $\dfrac{R}{S} = \dfrac{l}{100-l}$, where $l$ is the balance length measured from the end where $R$ is connected. Shunting $S$ with an external resistor reduces its effective resistance; since $R$ is unchanged, the ratio $R/S$ increases, which forces the balance length $l$ to increase (the null point moves toward the $S$ side) to keep the bridge balanced.

    Step-by-Step Solution

    1. Initial balance: $l = 20\ \mathrm{cm}$, so $\dfrac{R}{S} = \dfrac{20}{100-20} = \dfrac{20}{80} = \dfrac14$, i.e. $S = 4R$.
    2. Shunting $S$ with $60\ \Omega$ makes the new resistance $S' = \dfrac{S\times60}{S+60} < S$, which increases the ratio $R/S'$, so the null point moves further from $R$'s end: $l' = 20 + 5 = 25\ \mathrm{cm}$.
    3. New balance condition: $\dfrac{R}{S'} = \dfrac{25}{100-25} = \dfrac{25}{75} = \dfrac13 \Rightarrow S' = 3R$.
    4. Also $S' = \dfrac{60(4R)}{4R+60} = \dfrac{240R}{4R+60}$. …

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