Q.A galvanometer of resistance $10\ \Omega$ that gives maximum (full-scale) deflection for a current of $1\ \text{mA}$ is to be converted into a multirange voltmeter reading $2\ \text{V}$, $20\ \text{V}$ and $200\ \text{V}$. Three resistors $R_1$, $R_2$ and $R_3$ are joined in series with the galvanometer, one after another. The $2\ \text{V}$ terminal is tapped just after $R_1$, the $20\ \text{V}$ terminal after the series pair $R_1+R_2$, and the $200\ \text{V}$ terminal after $R_1+R_2+R_3$; each range terminal together with the common galvanometer terminal forms the two leads of the voltmeter for that range. Find $R_1$, $R_2$ and $R_3$.

Showing the 12 most recent of 14 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A galvanometer of resistance $50 \Omega$ is having 30 divisions and a current sensitivity $10 \mathrm{~mA} /$ div. What should be the shunt resistance so that it can be converted into an ammeter of range 10 A ? (A) $0.76 \Omega$ (B) $2.5 \Omega$ (C) $1.55 \Omega$ (D) $3.1 \Omega$

   ›Reveal solutionSolution

   The key idea is that a shunt resistor bypasses most of the current around the galvanometer. The full-scale deflection current of the galvanometer is 0.3 A, and to extend the range to 10 A, the shunt must carry the remaining 9.7 A. Using the parallel voltage equality, the required shunt resistance is approximately 0.0155 Ω, which corresponds to option (C).

   ---

   Concept and Intuition

   A galvanometer is a sensitive current-measuring device that deflects fully when a small current (here, 0.3 A) passes through it. To measure larger currents (like 10 A), we connect a shunt resistor in parallel with the galvanometer. The shunt "steals" most of the current, allowing only the safe, small current through the galvanometer. The trick is that the voltage across the galvanometer and the shunt must be equal (since they are in parallel). This voltage equality gives us the shunt resistance directly.

   ---

   Step-by-Step Solution

   1. Find the full-scale deflection current of the galvanometer. The galvanometer has 30 divisions and a sensitivity of $10 \, \text{mA/div}$. So the current for full-scale deflection is:

   $I_g = 30 \times 10 \, \text{mA} = 300 \, \text{mA} = 0.3 \, \text{A}.$

   2. Determine the current that must flow through the shunt. The ammeter's total range is $I = 10 \, \text{A}$. When the galvanometer shows full deflection (0.3 A through it), the remaining current goes through the shunt:

   $I_s = I - I_g = 10 - 0.3 = 9.7 \, \text{A}.$

   3. Apply the parallel voltage condition. In a parallel circuit, voltage across the galvanometer equals voltage across the shunt: $$$$

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.A $50 \Omega$ galvanometer is shunted by a resistance of $S \Omega$. If $8 \%$ of total current passes through the galvanometer, the value of $S$ is: (A) $4 \Omega$ (B) $4.35 \Omega$ (C) $3.9 \Omega$ (D) $3.35 \Omega$

   ›Reveal solutionSolution

   The shunt resistor is chosen so that only 8% of the total current flows through the galvanometer. Using the current-divider rule for parallel resistors, the shunt resistance is found to be approximately 4.35 Ω, which corresponds to option (B).

   Concept & Intuition

   A galvanometer is a sensitive current-measuring device that can only handle a small current. To measure larger currents, we connect a low-resistance shunt in parallel with it. The shunt diverts most of the current away from the galvanometer.

   The key idea: In a parallel circuit, the current divides inversely with resistance. If only 8% of the total current goes through the galvanometer, then 92% goes through the shunt. The ratio of currents equals the inverse ratio of resistances.

   Step-by-step solution

   1. Define the currents

      Let $I$ be the total current entering the parallel combination.

      Current through galvanometer: $I_g = 8\% \text{ of } I = 0.08 I$

      Current through shunt: $I_s = I - I_g = 0.92 I$

   2. Apply the parallel current-divider rule

      For two resistors in parallel, the current through one branch is given by:

   $I_g = I \times \frac{S}{G + S}$

   where $G = 50\ \Omega$ is the galvanometer resistance and $S$ is the shunt resistance.

   (This formula works because the voltage across both is the same: $I_g G = I_s S$.)

   3. Substitute the known values

   $0.08 I = I \times \frac{S}{50 + S}$

   Cancel $I$ (assuming $I \neq 0$):

   $0.08 = \frac{S}{50 + S}$

   4. Solve for $S$ Multiply both sides by $50 + S$: …

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3. KCET 2025Set D-41 markMCQ
   Q.If $r$ and $r'$ denotes the angles inside the prism having angle of prism $50^\circ$ considering that during interval of time from $t = 0$ to $t = t$, $r$ varies with time as $r = 10^\circ + t^2$. During the time $r'$ will vary with time as (A) $40^\circ + t^2$ (B) $50^\circ - t^2$ (C) $50^\circ + t^2$ (D) $40^\circ - t^2$

   ›Reveal solutionSolution

   Use the prism relation $r + r' = A$ — the two internal refraction angles must always sum to the prism angle — and substitute the given time-dependence of $r$.

   Step 1 — The prism geometry relation.

   Consider a ray entering face AB of a prism of apex angle $A$ and leaving through face AC. Let $r$ be the angle of refraction at the first face and $r'$ the angle of incidence on the second face (both measured inside the glass, from the respective normals).

   In the quadrilateral formed by the two faces and the two normals, the normals meet the faces at $90^\circ$ each, so the angle between the normals is $180^\circ - A$. In the triangle formed by the ray inside the prism and the two normals:

   $r + r' + (180^\circ - A) = 180^\circ$

   $\boxed{\ r + r' = A\ }$

   This is a purely geometric result — it holds at every instant, whatever the refractive index or the angle of incidence.

   Step 2 — Insert the given data.

   The prism angle is $A = 50^\circ$, and we are told that during $t = 0$ to $t = t$,

   $r = 10^\circ + t^2$

   Step 3 — Solve for $r'$.

   $r' = A - r = 50^\circ - (10^\circ + t^2)$

   $r' = 40^\circ - t^2$

   Step 4 — Physical check.

   At $t = 0$: $r = 10^\circ$ and $r' = 40^\circ$; their sum is $50^\circ = A$ ✓. …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.A galvanometer of $50 \Omega$ resistance is converted into an ammeter using a shunt resistance of $10 \Omega$. If the same resistance is used to convert the same galvanometer in to a voltmeter, what would be the ratio of the resistance of the ammeter to the resistance of the voltmeter? (A) $5: 1$ (B) $5: 36$ (C) $5: 6$ (D) $1: 5$

   ›Reveal solutionSolution

   The key idea is to compute the effective resistance of the galvanometer-shunt combination when used as an ammeter (parallel) and as a voltmeter (series), then take their ratio. The result is $5:36$.

   Concept & Intuition

   A galvanometer is a sensitive current-measuring device with a fixed internal resistance $R_g$. To convert it into an ammeter, we connect a small shunt resistance $R_s$ in parallel, so most current bypasses the coil. The ammeter’s effective resistance is the parallel combination of $R_g$ and $R_s$, which is very low.

   To convert it into a voltmeter, we connect a large series resistance $R_s$ (the same value here) so that most voltage drops across it. The voltmeter’s effective resistance is the series combination of $R_g$ and $R_s$, which is high.

   The problem asks for the ratio of these two effective resistances.

   Step-by-step solution

   1. Identify given values

      Galvanometer resistance: $R_g = 50\ \Omega$

      Shunt/series resistance: $R_s = 10\ \Omega$

   2. Effective resistance of the ammeter

      Ammeter: shunt in parallel with galvanometer.

   $R_A = \frac{R_g R_s}{R_g + R_s} = \frac{50 \times 10}{50 + 10} = \frac{500}{60} = \frac{25}{3}\ \Omega$

   3. Effective resistance of the voltmeter Voltmeter: series resistance added to galvanometer.

   $R_V = R_g + R_s = 50 + 10 = 60\ \Omega$

   4. Compute the ratio …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A galvanometer having a resistance of $50 \Omega$ is shunted by a wire of resistance $10 \Omega$. If the total current is 2 A , the current passing through the shunt is: (A) $\frac{3}{5} \mathrm{~A}$ (B) $\frac{2}{5} \mathrm{~A}$ (C) $\frac{5}{6} \mathrm{~A}$ (D) $\frac{5}{3} \mathrm{~A}$

   ›Reveal solutionSolution

   The shunt and galvanometer are in parallel, so they share the total current in inverse proportion to their resistances. Using the current divider rule, the current through the shunt is $\frac{5}{3}\,\text{A}$, which corresponds to option (D).

   Concept & Intuition

   A shunt is a low-resistance path placed in parallel with a galvanometer to divert most of the current away from the delicate meter. Because the two branches are in parallel, the voltage across each is the same. By Ohm’s law, the current in each branch is inversely proportional to its resistance — the smaller resistance carries the larger current. This is the classic current divider principle.

   Step-by-step solution

   1. Identify the parallel branches

      The galvanometer resistance is $R_g = 50\,\Omega$ and the shunt resistance is $R_s = 10\,\Omega$. They are connected in parallel, so the voltage across both is identical: $V = I_g R_g = I_s R_s$.

   2. Apply the current divider relation

      For two resistors in parallel, the current through one branch is the total current multiplied by the opposite branch resistance divided by the sum of the resistances. Specifically, the current through the shunt is:

   $I_s = I_{\text{total}} \times \frac{R_g}{R_g + R_s}$

   Why? Because the total current splits such that the larger resistance gets the smaller fraction — and here the shunt gets the fraction $\frac{R_g}{R_g+R_s}$.

   3. Plug in the numbers Total current $I_{\text{total}} = 2\,\text{A}$, $R_g = 50\,\Omega$, $R_s = 10\,\Omega$:

   $I_s = 2 \times \frac{50}{50 + 10} = 2 \times \frac{50}{60} = 2 \times \frac{5}{6} = \frac{10}{6} = \frac{5}{3}\,\text{A}.$

   4. Check consistency …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.A galvanometer of resistance $50 \Omega$ is connected to a battery of 4 V along with a resistance of $3950 \Omega$ in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 10 divisions, the resistance in series should be equal to: (A) 8950 ohm (B) 11950 ohm (C) 7000 ohm (D) 6000 ohm

   ›Reveal solutionSolution

   The key idea is that the galvanometer’s deflection is proportional to the current through it. By increasing the series resistance, we reduce the current to one‑third of its original value, which gives the new series resistance as 11950 Ω. The correct option is (B).

   ---

   We start with a galvanometer that shows a full‑scale deflection of 30 divisions when a certain current flows through it. The deflection is directly proportional to the current (assuming linear scale). So if we want the deflection to drop to 10 divisions (one‑third of 30), we need the current to become one‑third of its original value.

   Why this approach works:

   The total circuit resistance determines the current from the battery. Initially, the total resistance is the galvanometer’s own resistance plus the external series resistor. To reduce the current to one‑third, we must increase the total resistance to three times its original value. That lets us solve for the new series resistor.

   ---

   1. Find the initial total resistance and current. The galvanometer resistance is $R_g = 50\ \Omega$. The initial series resistance is $R_1 = 3950\ \Omega$. Total initial resistance:

   $R_{\text{total, initial}} = R_g + R_1 = 50 + 3950 = 4000\ \Omega.$

   Battery voltage $V = 4\ \text{V}$.

   Initial current:

   $I_1 = \frac{V}{R_{\text{total, initial}}} = \frac{4}{4000} = 0.001\ \text{A} = 1\ \text{mA}.$

   This current produces 30 divisions.

   2. Determine the current needed for 10 divisions. Since deflection ∝ current, for 10 divisions we need:

   $I_2 = \frac{10}{30} \times I_1 = \frac{1}{3} \times 1\ \text{mA} = \frac{1}{3}\ \text{mA}.$

   3. Find the total resistance required for this new current. Using Ohm’s law: $R_{\text{total, new}} = \frac{V}{I_2} = \frac{4\ \text{V}}{(1/3)\ \text{mA}} = \frac{4}{1/3 \times 10^{-3}} = 4 \times 3 \times 10^3 = 12000\ \Omega.$ …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.To increase the current sensitivity of a moving coil galvanometer by $25 \%$, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does the voltage sensitivity change? (A) Decreases by $62.5 \%$ (B) Decreases by $37.5 \%$ (C) Increases by $37.5 \%$ (D) Increases by $62.5 \%$

   ›Reveal solutionSolution

   Current sensitivity increases by 25% and resistance doubles, so voltage sensitivity (current sensitivity ÷ resistance) changes by a factor of 0.625, meaning it decreases by 37.5%. The correct option is (B).

   Concept & Intuition

   A moving coil galvanometer’s current sensitivity is the deflection per unit current: $S_i = \frac{\theta}{I}$. Its voltage sensitivity is the deflection per unit voltage: $S_v = \frac{\theta}{V}$. Since $V = IR$ (by Ohm’s law for the galvanometer circuit), we have

   $S_v = \frac{\theta}{IR} = \frac{S_i}{R}.$

   So voltage sensitivity is directly proportional to current sensitivity and inversely proportional to resistance. If we change both $S_i$ and $R$, the factor change in $S_v$ is simply the ratio of the factor change in $S_i$ to the factor change in $R$.

   Step-by-step reasoning

   1. Interpret the given changes

      Current sensitivity increases by 25%, meaning it becomes $1 + 0.25 = 1.25$ times its original value.

      Resistance becomes twice its initial value, so the factor for $R$ is $2$.

   2. Write the relation for voltage sensitivity

   $S_v = \frac{S_i}{R}$

   Let initial values be $S_{i0}$, $R_0$, and $S_{v0}$. Then

   $S_{v0} = \frac{S_{i0}}{R_0}.$

   3. Compute the new voltage sensitivity New values:

   $S_i' = 1.25\, S_{i0}, \quad R' = 2\, R_0.$

   Hence

   $S_v' = \frac{S_i'}{R'} = \frac{1.25\, S_{i0}}{2\, R_0} = 0.625 \cdot \frac{S_{i0}}{R_0} = 0.625\, S_{v0}.$

   4. Interpret the factor …

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8. COMEDK 2024Set 2024-A1 markMCQ
   Q.Two identical moving coil galvanometers have $10 \Omega$ resistance and full-scale deflection at $2 \mu \mathrm{A}$ current. One of them is converted into a voltmeter of range $10 \mathrm{~mV}$ and the other into an ammeter of range $1 \mathrm{~mA}$ using appropriate resistors. The ratio of resistance of the converted voltmeter to that of the ammeter is (A) 5000 : 1 (B) 2500 : 1 (C) 250000 : 1 (D) 25000 : 1

   ›Reveal solutionSolution

   The converted voltmeter has resistance $5000\,\Omega$ and the ammeter about $0.02\,\Omega$, giving a ratio of $250000:1$.

   Given $R_g=10\,\Omega$, full-scale current $I_g=2\,\mu A$.

   Voltmeter (range $10$ mV):

   $R_V=\frac{V}{I_g}=\frac{10\times10^{-3}}{2\times10^{-6}}=5000\ \Omega.$

   Ammeter (range $1$ mA): shunt $S$ carries $I-I_g$ while the coil carries $I_g$: …

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9. COMEDK 2024Set 2024-E1 markMCQ
   Q.A voltmeter of resistance $1000 \Omega .0 .5 \mathrm{~V} /$ div is to be converted into a voltmeter to make it to read $=1 \mathrm{~V} / \mathrm{div}$. The value of high resistance to be connected in series with it is (A) $6000 \Omega$ (B) $5000 \Omega$ (C) $4000 \Omega$ (D) $1000 \Omega$

   ›Reveal solutionSolution

   Doubling the volts-per-division needs total resistance $2000\,\Omega$, so add $1000\,\Omega$ in series.

   The meter's current per division is fixed:

   $i=\frac{0.5\text{ V/div}}{1000\,\Omega}=5\times10^{-4}\text{ A/div}$

   To read $1$ V/div with the same $i$, the required total resistance is …

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10. KCET 2023Set A-31 markMCQ
    Q.Total impedance of a series LCR circuit varies with angular frequency of the AC source connected to it as shown in the graph. The quality factor $Q$ of the series LCR circuit is (A) 2.5 (B) 5 (C) 1 (D) 0.4

    ›Reveal solutionSolution

    Recognise the $\sqrt2\,Z_{\min}$ crossings as the half-power (bandwidth) frequencies, then use $Q = \dfrac{\omega_0}{\Delta\omega}$.

    1. Why $\sqrt{2}\,Z_{\min}$ marks the half-power points

    For a series LCR circuit,

    $Z(\omega) = \sqrt{R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}}$

    At resonance the reactances cancel ($\omega_0 L = 1/\omega_0 C$), leaving the minimum impedance

    $Z_{\min} = R \qquad\text{at}\qquad \omega_0 = \frac{1}{\sqrt{LC}}$

    The graph shows this minimum at $\omega_0 = 500\ \mathrm{rad\,s^{-1}}$.

    The half-power points $\omega_1,\omega_2$ are where the power dissipated falls to half its resonant value. Since $P = \dfrac{V_{\text{rms}}^2 R}{Z^2}$, halving $P$ means $Z^{2} = 2R^{2}$, i.e.

    $Z = \sqrt{2}\,R = \sqrt{2}\,Z_{\min}$

    That is precisely the upper dashed line on the graph. So the two frequencies where the curve crosses it,

    $\omega_1 = 400\ \mathrm{rad\,s^{-1}}, \qquad \omega_2 = 600\ \mathrm{rad\,s^{-1}},$

    are the half-power frequencies. (Equivalently, the current there is $I_{\max}/\sqrt2$.)

    2. Bandwidth

    $\Delta\omega = \omega_2 - \omega_1 = 600 - 400 = 200\ \mathrm{rad\,s^{-1}}$

    3. Quality factor

    $Q = \frac{\omega_0}{\Delta\omega} = \frac{500}{200} = 2.5$ …

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11. COMEDK 2023Set 2023-E1 markMCQ
    Q.The current sensitivity of a galvanometer having 20 divisions is $10 \mu \mathrm{A} /$ div. If the resistance of the galvanometer is $100 \Omega$ then the value of the resistance to be used to convert this galvanometer in to an voltmeter to read up to $1 \mathrm{~V}$ is : (A) $4 \times 10^{-6} \Omega$ in series with the galvanometer. (B) $4900 \Omega$ in parallel with the galvanometer. (C) $4 \times 10^{-3} \Omega$ in series with the galvanometer. (D) $4900 \Omega$ in series with the galvanometer.

    ›Reveal solutionSolution

    So a 4900 ohm resistance is connected IN SERIES with the galvanometer. (A parallel resistance would make an ammeter, so (B) is wrong; the tiny values in (A)/(C) are shunt-like and far too small.)

    Concept: converting a galvanometer into a voltmeter - connect a HIGH resistance R in SERIES, chosen so that the full-scale current Ig produces the desired full-scale voltage V:

    V = Ig (G + R) => R = V/Ig - G.

    Step 1 - full-scale current:

    Ig = (current sensitivity) x (number of divisions) = 10 uA/div x 20 div = 200 uA = 2 x 10^-4 A.

    Step 2 - series resistance for V = 1 V, G = 100 ohm:

    R = 1 / (2 x 10^-4) - 100 = 5000 - 100 = 4900 ohm. …

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12. KCET 2022Set B-31 markMCQ
    Q.When a metal conductor connected to left gap of a meter bridge is heated, the balancing point (A) Remains unchanged (B) Shifts to the center (C) Shifts towards right (D) Shifts towards left

    ›Reveal solutionSolution

    Metals have a positive temperature coefficient, so heating raises the left-gap resistance; the metre-bridge balance condition then demands a longer left-hand length, i.e. the null point slides right.

    Step 1 — The balance condition.

    A metre bridge is a Wheatstone bridge whose two lower arms are the two parts of a uniform 1 m wire. With $R$ in the left gap and $S$ in the right gap, and the jockey balanced at a distance $l$ (in cm) from the left end, the resistances of the wire segments are proportional to their lengths, so

    $\frac{R}{S}=\frac{l}{100-l}.$

    Step 2 — What heating does to a metal.

    For a metallic conductor,

    $R_T=R_0\left[1+\alpha\,\Delta T\right],\qquad \alpha>0 .$

    Heating makes the lattice ions vibrate more vigorously, so conduction electrons are scattered more often; the relaxation time $\tau$ falls and, since $\rho=\dfrac{m}{ne^{2}\tau}$, the resistivity — and hence $R$ — increases. (This is the key difference from a semiconductor, where more carriers are freed and $R$ falls.)

    Step 3 — Track the balance point.

    $S$ (right gap) is untouched, so the ratio $R/S$ increases. Therefore $\dfrac{l}{100-l}$ must increase. Since this fraction is a strictly increasing function of $l$, the new balancing length $l$ must be larger than before — the null point is found further from the left end, i.e. it shifts towards the right. …

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