Q.Four resistors, each of resistance $R$, and a key $K$ are connected as shown in the figure. The equivalent resistance between points A and B when key $K$ is open will be: (A) $4R$ (B) $\infty$ (C) $\dfrac{R}{4}$ (D) $\dfrac{4R}{3}$

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.The equivalent resistance of resistors in parallel combination (A) decreases (B) increases (C) remains same (D) none of these

   ›Reveal solutionSolution

   Adding resistors in parallel opens up more current paths, so the equivalent resistance falls below the smallest branch resistance.

   For resistors in parallel the reciprocals add:

   $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

   Because $\frac{1}{R_{eq}}$ is larger than any single $\frac{1}{R_i}$, the equivalent resistance $R_{eq}$ is smaller than every one of the individual resistances. Physically, more parallel paths give the current more routes to flow, lowering the overall opposition.

   ✓Final answer

   (A) decreases.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.There are n resistors each of resistance R. When they are connected in parallel, the equivalent resistance is x. If they are connected in series, the equivalent resistance will be

   (a) x/n^2

   (b) n^2 x

   (c) x/n

   (d) nx

   ›Reveal solutionSolution

   If $n$ equal resistors $R$ give parallel resistance $x$, their series resistance is $n^2x$.

   For $n$ identical resistors, each of resistance $R$, connected in parallel:

   $\dfrac{1}{R_{\text{parallel}}} = \dfrac{n}{R} \ \Rightarrow\ R_{\text{parallel}} = \dfrac{R}{n} = x \ \Rightarrow\ R = nx$

   When the same $n$ resistors are connected in series:

   $R_{\text{series}} = nR = n(nx) = n^2x$

   ✓Final answer

   (b) $n^2x$.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the ammeter in the given circuit reads 2 amperes, the resistance R is

   (a) 1 ohm

   (b) 2 ohms

   (c) 3 ohms

   (d) 4 ohms

   ›Reveal solutionSolution

   The 3 Ω and 6 Ω resistors in parallel give 2 Ω; using the total emf and ammeter current gives $R = 1\ \Omega$.

   The 3 Ω and 6 Ω resistors are in parallel:

   $R_p = \dfrac{3\times6}{3+6} = \dfrac{18}{9} = 2\ \Omega$

   This parallel combination is in series with $R$, an ammeter (ideal, zero resistance) and a 6 V battery with zero internal resistance, all in one loop. The ammeter reads $I = 2$ A, so by Ohm's law for the whole loop:

   $\varepsilon = I(R_p + R) \ \Rightarrow\ 6 = 2(2+R) \ \Rightarrow\ 2+R = 3 \ \Rightarrow\ R = 1\ \Omega$

   ✓Final answer

   (a) 1 ohm.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.If a uniform wire of resistance 16 ohm is cut into four equal parts and attached in parallel combination, the equivalent resistance of the combination is

   (a) 1 ohm

   (b) 4 ohm

   (c) 1/4 ohm

   (d) 1/16 ohm

   ›Reveal solutionSolution

   Cutting a uniform wire into N equal pieces divides each piece's resistance by N; connecting those pieces in parallel divides the equivalent resistance by N again.

   Resistance of a uniform wire is proportional to its length: R = rho*l/A. If a wire of resistance R = 16 ohm is cut into 4 EQUAL parts, each part is 1/4 the original length, so each part has resistance

   r = R/4 = 16/4 = 4 ohm

   Now these four 4 ohm resistors are connected in PARALLEL. For resistors in parallel,

   1/R_eq = 1/r + 1/r + 1/r + 1/r = 4/r = 4/4 = 1

   So R_eq = 1 ohm.

   ✓Final answer

   (a) 1 ohm.

5. CBSE 2025Set 55/5/11 markMCQ
   Q.Four resistors, each of resistance $R$, and a key $K$ are connected as shown in the figure. The equivalent resistance between points A and B when key $K$ is open will be: (A) $4R$ (B) $\infty$ (C) $\dfrac{R}{4}$ (D) $\dfrac{4R}{3}$

   ›Reveal solutionSolution

   When key $K$ is open, the circuit reduces to a single resistor $R$ from A to the central node O, followed by three resistors $R$ in parallel from O to the ring (which is all one node including B). The equivalent resistance is $R + \frac{R}{3} = \frac{4R}{3}$.

   

   The key insight here is that the conducting ring is a perfect conductor — it has zero resistance. That means every point on the ring is at the same electrical potential. So the top (T), bottom (Bot), left (L), and terminal B are all the same node. This is the heart of the problem.

   When $K$ is open, the branch containing $K$ carries no current — it's just an open circuit. So current from A can only go through the resistor $R$ to the central node O. From O, there are three paths: through the three resistors $R$ to T, Bot, and L — but all three of those points are connected to the ring, which is node B. So those three resistors are in parallel between O and B.

   Let's work through it step by step.

   1. Identify the nodes with the ring closed.

      The ring is a zero-resistance wire. Points T, Bot, L, and B are all shorted together. So they form a single electrical node — call it node B for convenience.

   2. Trace the path from A to B with $K$ open.

      Starting at A, the only way to reach B is:

      A → resistor $R$ → central node O → then through any of the three resistors $R$ to the ring (node B).

      The $K$ branch is open, so no current flows through it.

   3. Find the equivalent resistance from O to B.

      From O, three resistors $R$ each go to the ring (node B). Since all three share the same two nodes (O and B), they are in parallel.

      The parallel combination of three equal resistors $R$ is:

   $R_{\text{parallel}} = \frac{R}{3}$

   4. Add the series resistor from A to O. The resistor from A to O is in series with the parallel combination from O to B. So the total equivalent resistance between A and B is:

   $R_{AB} = R + \frac{R}{3} = \frac{4R}{3}$

   Watch out

   A common mistake is to think the ring introduces extra series paths or that the three resistors from O to the ring are in series with each other. But because the ring is a perfect conductor, all three connect to the same node — they are in parallel, not series.

   Tip

   Whenever you see a conducting ring or wire connecting multiple points in a circuit, treat all those points as a single node. This instantly simplifies the topology.

   ✓Final answer

   The equivalent resistance between A and B when key $K$ is open is $\boxed{\frac{4R}{3}}$, which corresponds to option (D).

6. CBSE 2025Set D1 markMCQ
   Q.In series circuit with resistors R1, R2 and R3, the current flowing through each resistor is (A) same (B) different (C) zero (D) divided proportionally to the value of resistance

   ›Reveal solutionSolution

   Current is the same through all resistors in series.

   In a series combination the resistors form a single unbranched path, so charge has nowhere else to go — the same current I flows through R₁, R₂ and R₃. What differs is the voltage drop across each (V₁ = IR₁, etc.), which is larger for a larger resistance. The equivalent resistance is R = R₁ + R₂ + R₃.

   ✓Final answer

   (A) same.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The ammeter reading in the adjoining circuit is

   (a) 1/5 A

   (b) 2/5 A

   (c) 3/5 A

   (d) 4/5 A

   ›Reveal solutionSolution

   This is a balanced Wheatstone-bridge-shaped network: the 10-ohm arm carries no current, so the circuit reduces to two equal 10-ohm paths in parallel; the ammeter (in one arm) reads half the total current.

   Label the left corner node L, right corner node R (battery terminals), top node T, and middle node M.

   • Path 1 (upper): L $\to$ T $\to$ R, resistances 5 $\Omega$ + 5 $\Omega$ = 10 $\Omega$
   • Path 2 (lower): L $\to$ M $\to$ R, resistances 5 $\Omega$ + 5 $\Omega$ = 10 $\Omega$
   • Bridge element: T $\to$ M, 10 $\Omega$

   Check the Wheatstone-bridge balance condition on the two arms meeting at L and R: $\dfrac{R_{LT}}{R_{TR}} = \dfrac{5}{5} = 1$ and $\dfrac{R_{LM}}{R_{MR}} = \dfrac{5}{5}=1$ — equal ratios, so the bridge is balanced and no current flows through the 10 $\Omega$ bridge arm (T-M).

   With the bridge arm carrying no current, the network is simply two 10 $\Omega$ paths (L-T-R and L-M-R) in parallel between the battery terminals:

   $R_{eq} = \dfrac{10 \times 10}{10+10} = 5\ \Omega$

   Total current from the 2 V battery: $I = \dfrac{V}{R_{eq}} = \dfrac{2}{5} = 0.4$ A

   Since the two paths have equal resistance (10 $\Omega$ each), this total current divides equally between them:

   $I_{each\ path} = \dfrac{0.4}{2} = 0.2\ \text{A} = \dfrac{1}{5}\ \text{A}$

   The ammeter sits in one arm of the left side (one of the two symmetric branches), so it reads this half-current.

   ✓Final answer

   (a) 1/5 A.

8. CBSE 2024Set A1 markMCQ
   Q.n equal resistors are first connected in series and then in parallel. The ratio of maximum and minimum resistances will be (A) 1/n (B) n (C) 1/n^2 (D) n^2

   ›Reveal solutionSolution

   R_series = nR (maximum), R_parallel = R/n (minimum) → ratio = nR ÷ (R/n) = n².

   For n equal resistors each of resistance R:

   • Series (maximum): $R_{max} = nR$.
   • Parallel (minimum): $R_{min} = \dfrac{R}{n}$.

   $\frac{R_{max}}{R_{min}} = \frac{nR}{R/n} = n^2$

   ✓Final answer

   (D) n².

9. CBSE 2021Set A1 markMCQ
   Q.Which of the following is the same in parallel connection of resistances? (A) Potential difference (B) Current (C) Both Potential difference and current (D) None of these

   ›Reveal solutionSolution

   In a parallel combination the potential difference across each resistor is the same; the current divides.

   When resistors are connected in parallel, both ends of every resistor are joined to the same pair of nodes. Therefore each resistor has the same potential difference V across it.

   The current, however, splits between the branches according to Ohm's law $I = V/R$ — a smaller resistance carries a larger current. So current is NOT the same; only the voltage is common. (In a series combination it is the reverse: current is common, voltage divides.)

   ✓Final answer

   (A) Potential difference.

10. CBSE 2020Set XS1 markMCQ
    Q.A wire of 24 ohm resistance is bent in the form of an equilateral triangle. The effective resistance between any two corners is:

    (i) $\dfrac{9}{2}$ ohm

    (ii) $\dfrac{16}{3}$ ohm

    (iii) 12 ohm

    (iv) 24 ohm

    ›Reveal solutionSolution

    Three 8 $\Omega$ sides; one side is parallel to the other two in series $\Rightarrow \frac{16}{3}\,\Omega$.

    Set-up. A 24 $\Omega$ wire bent into an equilateral triangle gives three equal sides of $24/3 = 8\,\Omega$ each.

    Between two corners. Current has two paths:

    • the direct side: $8\,\Omega$
    • the other two sides in series: $8+8 = 16\,\Omega$

    These two paths are in parallel:

    $R_{eff}=\frac{8\times16}{8+16}=\frac{128}{24}=\frac{16}{3}\,\Omega\approx5.33\,\Omega.$

    ✓Final answer

    (ii) $\dfrac{16}{3}\,\Omega$.

11. CBSE 2020Set ANNUAL1 markMCQ
    Q.If a uniform wire of resistance 16 ohm is cut into four equal parts and attached in parallel combination, the equivalent resistance is

    (a) 1 ohm

    (b) 4 ohm

    (c) 1/4 ohm

    (d) 1/16 ohm

    ›Reveal solutionSolution

    Cutting a wire into n equal parts divides its resistance by n; reconnecting those n parts in parallel divides the resulting resistance by n again, so the net effect is R divided by n squared.

    A uniform wire's resistance is R = (rho L)/A. Cutting it into 4 equal-length parts gives each part a resistance of 16/4 = 4 ohm (since resistance is proportional to length, for the same wire material and cross-section).

    These four 4 ohm resistors are now connected in parallel. For resistors in parallel:

    1/R_eq = 1/4 + 1/4 + 1/4 + 1/4 = 4/4 = 1

    So R_eq = 1 ohm.

    ✓Final answer

    (a) 1 ohm.

12. CBSE 2020Set OC1 markMCQ
    Q.The total resistance of two resistors when connected in series is $9\ \Omega$ and when connected in parallel is $2\ \Omega$. The values of two resistances are

    (a) $4\ \Omega$ and $5\ \Omega$

    (b) $2\ \Omega$ and $7\ \Omega$

    (c) $3\ \Omega$ and $6\ \Omega$

    (d) $1\ \Omega$ and $8\ \Omega$

    ›Reveal solutionSolution

    Series sum and parallel product-over-sum give two equations in the two unknown resistances, solvable as a quadratic.

    Step 1 — Two equations

    $R_1+R_2 = 9 \qquad \frac{R_1R_2}{R_1+R_2} = 2 \implies R_1R_2 = 18$

    Step 2 — Quadratic in $x=R_1$ (with $R_2=9-x$)

    $x(9-x) = 18 \implies x^2-9x+18=0$

    $x = \frac{9\pm\sqrt{81-72}}{2} = \frac{9\pm3}{2} = 6\ \text{or}\ 3$

    ✓Final answer

    The two resistances are $3\ \Omega$ and $6\ \Omega$ — option (c).