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. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A galvanometer has 30 divisions on its scale. It requires a potential difference of 30 mV for a current of 1 mA through its coil, which produces a deflection of one division. The shunt required to convert it into an ammeter of range 0 - 3A is (A) $\dfrac{1}{33}\Omega$ (B) $\dfrac{20}{99}\Omega$ (C) $3.3\ \Omega$ (D) $\dfrac{10}{33}\Omega$

   ›Reveal solutionSolution

   Find the galvanometer's own resistance and full-scale current from the given division data, then apply the shunt formula $S=I_gG/I_s$ to get $S=\dfrac{10}{33}\Omega$.

   Concept and Intuition

   A galvanometer is converted into an ammeter by placing a low-resistance shunt in parallel with it, so that most of the current bypasses the delicate coil. At full-scale deflection, the galvanometer carries its maximum safe current $I_g$ while the shunt carries the rest of the total current $I$; since they're in parallel, the voltage across both must be equal, which fixes the shunt resistance.

   Step-by-Step Solution

   1. One division needs 1 mA through the coil and 30 mV across it, so the galvanometer's resistance is $G = \dfrac{30\text{ mV}}{1\text{ mA}} = 30\ \Omega$ (this ratio is the same at any deflection, since it's linear).
   2. Full-scale deflection (30 divisions) needs a coil current of $I_g = 30\times1\text{ mA} = 30\text{ mA} = 0.03$ A.
   3. For a 0–3 A ammeter, at full-scale the total current is 3 A, so the shunt must carry $I_s = 3 - 0.03 = 2.97$ A. …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The current and voltage sensitivities of a moving coil galvanometer are 80 divisions per mA and 2 divisions per mV respectively. If the galvanometer has 100 divisions, then the resistance to be connected in series to the galvanometer to convert it into a voltmeter which can measure a maximum potential difference of 5 V is (A) $4040\ \Omega$ (B) $3960\ \Omega$ (C) $5040\ \Omega$ (D) $4960\ \Omega$

   ›Reveal solutionSolution

   Converting a galvanometer to a voltmeter needs a large resistor in series so that at full-scale deflection the total voltage across (G+R) equals the desired range. We first extract $I_g$ and $G$ from the given sensitivities. Answer: $3960\ \Omega$.

   Concept and Intuition

   Current sensitivity tells us how many divisions deflect per unit current — inverting it at full scale gives the full-scale current $I_g$. Voltage sensitivity similarly gives the full-scale voltage across the bare galvanometer coil, from which $G = V_{fs}/I_g$. To extend the range to a voltmeter of $V_{max}$, we need $I_g(G+R) = V_{max}$.

   Step-by-Step Solution

   1. Current sensitivity $= 80$ div/mA, full scale $=100$ div $\Rightarrow I_g = \dfrac{100}{80} = 1.25$ mA $= 1.25\times10^{-3}$ A.
   2. Voltage sensitivity $= 2$ div/mV, full scale $=100$ div $\Rightarrow V_{fs}$ (across galvanometer alone) $= \dfrac{100}{2} = 50$ mV $= 0.05$ V.
   3. Galvanometer resistance: $G = \dfrac{V_{fs}}{I_g} = \dfrac{0.05}{1.25\times10^{-3}} = 40\ \Omega$. …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per mA and Voltage sensitivity is 2 divisions per millivolt. In order to read each division 1 volt, the resistance in ohms to be connected in series is (A) 9995 (B) 995 (C) 95 (D) 99995

   ›Reveal solutionSolution

   Tests converting a galvanometer's current and voltage sensitivities into its internal resistance and full-scale current, then finding the series resistance needed to convert it into a voltmeter of a specified range.

   Concept and Intuition

   "Sensitivity" here (divisions per mA, divisions per mV) tells us how the galvanometer coil itself responds — from these two numbers we can back out both its full-scale current (deflection capacity) and its own internal (coil) resistance, since voltage sensitivity implicitly assumes current is flowing only through the coil resistance. Once we know the full-scale current, converting the meter to read any desired full-scale voltage is just Ohm's law: find the total resistance needed for that current to flow at that voltage, and the extra beyond the coil's own resistance is the series resistor to add.

   Step-by-Step Solution

   1. Full-scale current $I_g$: sensitivity is $10$ divisions per mA, and full scale is $150$ divisions, so $I_g = \dfrac{150}{10} = 15\,mA = 0.015\,A$.
   2. Full-scale voltage across the bare galvanometer coil $V_g$: sensitivity is $2$ divisions per mV, so $V_g = \dfrac{150}{2} = 75\,mV = 0.075\,V$.
   3. Galvanometer (coil) resistance: $G = \dfrac{V_g}{I_g} = \dfrac{0.075}{0.015} = 5\,\Omega$. …

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4. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.In a galvanometer, 15% of total current in the circuit passes through it. If the resistance of the galvanometer is $G$, then the Shunt resistance that is connected to galvanometer is (A) $\dfrac{17}{3}G$ (B) $\dfrac{16}{3}G$ (C) $\dfrac{5}{17}G$ (D) $\dfrac{3}{17}G$

   ›Reveal solutionSolution

   Equal voltage across galvanometer and shunt (they're in parallel) with a 15%/85% current split gives $S = \tfrac{3}{17}G$.

   Concept and Intuition

   A shunt is a low resistance connected in parallel with the galvanometer to divert most of the current around it (protecting the sensitive coil while still allowing current measurement). Since they share the same voltage (parallel branches), the current in each branch is inversely proportional to its resistance.

   Step-by-Step Solution

   1. Let total current be $I$. Galvanometer current $I_g = 0.15I$; shunt current $I_s = I - I_g = 0.85I$.
   2. Parallel branches share the same potential difference: $I_g G = I_s S$. …

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5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.A maximum current of 0.5 mA can pass through a galvanometer of resistance 15 Ω. The resistance to be connected in series to the galvanometer to convert it into a voltmeter of range $0-10$ V is (A) 9985 Ω (B) 20015 Ω (C) 20000 Ω (D) 19985 Ω

   ›Reveal solutionSolution

   Converting a galvanometer to a voltmeter needs a large series resistance so that full-scale current flows exactly at full-scale voltage; here that resistance is 19985 Ω.

   Concept and Intuition

   A voltmeter must draw only the galvanometer's maximum safe current $I_g$ even when the full voltage range $V$ is applied across it. Placing a large resistance $R$ in series with the galvanometer (total resistance $R+R_g$) limits the current to exactly $I_g$ at $V$, converting a sensitive low-resistance current-measuring device into a high-resistance voltage-measuring device.

   Step-by-Step Solution

   1. For the voltmeter to read full range $V=10$ V at the galvanometer's maximum current $I_g = 0.5\ \mathrm{mA} = 5\times10^{-4}\ \mathrm A$: $V = I_g(R+R_g)$.
   2. Solve for $R$: $R = \dfrac{V}{I_g} - R_g$. …

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6. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.A galvanometer having 30 divisions has a current sensitivity of $0.0625\ \dfrac{\mathrm{div}}{\mu A}$. If it is converted into a voltmeter to read a maximum of 6 V, then the resistance of that voltmeter is (A) $7.5\ \mathrm{k}\Omega$ (B) $12.5\ \mathrm{k}\Omega$ (C) $6\ \mathrm{k}\Omega$ (D) $5\ \mathrm{k}\Omega$

   ›Reveal solutionSolution

   This tests converting a galvanometer's full-scale deflection current into the voltmeter resistance needed for a given full-scale voltage; the answer is (B) $12.5\ \mathrm{k}\Omega$.

   Concept and Intuition

   A galvanometer deflects fully when a certain current $I_g$ flows through it; "current sensitivity" tells us how many divisions of deflection occur per unit current, so dividing the total number of divisions by the sensitivity gives $I_g$. To use the galvanometer as a voltmeter reading up to some maximum voltage $V$, we need the total resistance of the voltmeter to be $R = V/I_g$, so that exactly $I_g$ flows at full-scale voltage $V$.

   Step-by-Step Solution

   1. Current sensitivity $= 0.0625\ \mathrm{div}/\mu A$, and full scale is 30 divisions.
   2. Full-scale current: $I_g = \dfrac{30\ \mathrm{div}}{0.0625\ \mathrm{div}/\mu A} = 480\ \mu A = 4.8\times10^{-4}\ \mathrm{A}$. …

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7. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.The value of shunt resistance, that allows only 10% of main current through the galvanometer of resistance 99 Ω is (A) 9 Ω (B) 4 Ω (C) 2 Ω (D) 11 Ω

   ›Reveal solutionSolution

   A shunt is a low resistance placed in parallel with the galvanometer to bypass most of the

   current; equal voltage across the parallel pair gives $S=11\,\Omega$ for the required 10%/90%

   split.

   Concept and Intuition

   A shunt diverts the bulk of the current away from the delicate galvanometer coil. Since the

   galvanometer (resistance G) and the shunt (resistance S) are connected in parallel, the voltage

   across them is identical, even though the currents through them differ.

   Step-by-Step Solution

   1. Let total (main) current be $I$. Given: current through galvanometer $I_g = 0.1I$, so current through the shunt $I_s = 0.9I$.
   2. Equal voltage across the parallel combination: $I_g G = I_s S$
   3. $(0.1I)(99) = (0.9I) S$ …

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8. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.Current sensitivities of two galvanometers $G_1$ and $G_2$ of resistances $100\ \Omega$ and $50\ \Omega$ are $10^{8}\ \mathrm{div/A}$ and $0.5\times10^{5}\ \mathrm{div/A}$ respectively. The galvanometer in which the voltage sensitivity is more is (A) Same in both galvanometers (B) More in $G_2$ (C) Zero (D) More in $G_1$

   ›Reveal solutionSolution

   Voltage sensitivity equals current sensitivity divided by galvanometer resistance; computing both shows $G_1$ has far greater voltage sensitivity.

   Concept and Intuition

   Current sensitivity $S_I$ tells you how many divisions deflect per ampere. Voltage sensitivity $S_V$ tells you how many divisions deflect per volt applied. Since $V = IG$ for a galvanometer of resistance $G$ carrying current $I$, a deflection produced by current $I$ corresponds to a voltage $V = IG$ across the coil, so $S_V = \dfrac{\text{deflection}}{V} = \dfrac{\text{deflection}}{IG} = \dfrac{S_I}{G}$.

   Step-by-Step Solution

   1. For $G_1$: $R_1 = 100\ \Omega$, $S_{I1}=10^8$ div/A.

   $S_{V1} = \frac{S_{I1}}{R_1} = \frac{10^8}{100} = 10^6\ \text{div/V}.$

   2. For $G_2$: $R_2 = 50\ \Omega$, $S_{I2}=0.5\times10^5 = 5\times10^4$ div/A.

   $S_{V2} = \frac{S_{I2}}{R_2} = \frac{5\times10^4}{50} = 1\times10^3\ \text{div/V}.$

   3. Compare: $S_{V1}=10^6 \gg S_{V2}=10^3$. …

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9. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.In a galvanometer, when the number of turns N becomes 2N and resistance R becomes doubled, then (A) current sensitivity remains unchanged and voltage sensitivity doubles. (B) current sensitivity and voltage sensitivity remain unchanged. (C) current sensitivity doubles and voltage sensitivity remain unchanged. (D) current sensitivity and voltage sensitivity doubles.

   ›Reveal solutionSolution

   Doubling both the turns and the coil resistance doubles the galvanometer's current sensitivity while leaving its voltage sensitivity unchanged.

   Concept and Intuition

   Current sensitivity measures deflection per unit current, $I_s = NBA/k$ — it grows directly with the number of turns $N$ (more turns means more torque per unit current). Voltage sensitivity measures deflection per unit voltage, $V_s = NBA/(kR) = I_s/R$ — it depends on both $N$ and the coil's own resistance $R$, since a given voltage drives less current through a higher-resistance coil.

   Step-by-Step Solution

   1. Current sensitivity: $I_s = \dfrac{NBA}{k}$. With $N\to 2N$: $I_s' = \dfrac{2NBA}{k}=2I_s$ — doubles. …

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10. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.A galvanometer has a coil of resistance 100 $\Omega$ showing a full scale deflection at 50 $\mu A$. The resistance that should be added to use it as an ammeter of range 10 mA is (A) $5\ \Omega$ (B) $5\times10^{-2}\ \Omega$ (C) $0.5\ \Omega$ (D) $1\ \Omega$

    ›Reveal solutionSolution

    The required shunt is $S=\dfrac{I_g G}{I-I_g}\approx 0.5\ \Omega$.

    Concept and Intuition

    A shunt of resistance $S$ in parallel with the coil ($G$) carries the excess current $I-I_g$ while the coil carries its full-scale $I_g$. Equal voltage across both gives $I_g G = (I-I_g)S$.

    Step-by-Step Solution

    1. Data: $G=100\ \Omega$, $I_g=50\ \mu A=50\times10^{-6}$ A, $I=10\ \text{mA}=10\times10^{-3}$ A.
    2. $S=\dfrac{I_g G}{I-I_g}=\dfrac{(50\times10^{-6})(100)}{10\times10^{-3}-50\times10^{-6}}$.
    3. Numerator $=5\times10^{-3}$; denominator $=9.95\times10^{-3}$.
    4. $S=\dfrac{5\times10^{-3}}{9.95\times10^{-3}}\approx 0.5025\ \Omega \approx 0.5\ \Omega$.

    Common Mistakes

    • Treating the added resistance as a series multiplier (that converts a galvanometer to a voltmeter, not an ammeter). …

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11. AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQ
    Q.Two galvanometers A and B require 3 mA and 5 mA respectively to produce same deflection of '$I_0$' divisions. Then, (A) A is more sensitive than B (B) B is more sensitive than A (C) A and B are equally sensitive (D) sensitiveness of B is $\dfrac{5}{3}$ times that of A

    ›Reveal solutionSolution

    The galvanometer needing less current for the same deflection is the more sensitive one — that's A.

    Concept and Intuition

    Galvanometer sensitivity is defined as the deflection produced per unit current, $S=\theta/I$ (or equivalently, $S\propto 1/I$ for a fixed target deflection). A highly sensitive instrument can detect/display the same deflection with a much smaller input current, since less current is needed to produce the same torque-based swing of the needle/coil.

    Step-by-Step Solution

    1. Sensitivity for a fixed deflection $I_0$: $S=\dfrac{I_0}{I_{required}}$ (in relative terms, smaller required current means higher sensitivity).
    2. Galvanometer A needs $I_A=3\,mA$ for deflection $I_0$; galvanometer B needs $I_B=5\,mA$ for the same deflection.
    3. Since $I_A<I_B$, A achieves the same deflection with less current, so A is the more sensitive galvanometer. …

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12. AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
    Q.A galvanometer of resistance $80\ \Omega$ is converted to an ammeter by a shunt resistance $r_s = 0.04\ \Omega$. Value of its resistance is ______ (A) $0.02\ \Omega$ (B) $0.04\ \Omega$ (C) $0.08\ \Omega$ (D) $0.06\ \Omega$

    ›Reveal solutionSolution

    The ammeter's net (parallel) resistance is $\dfrac{G r_s}{G+r_s}\approx0.04\ \Omega$, essentially equal to the tiny shunt resistance since $G\gg r_s$.

    Concept and Intuition

    Converting a sensitive galvanometer into an ammeter means placing a low-resistance shunt in parallel with it, so that most of the current bypasses the delicate coil through the shunt, and only a small, safe fraction flows through the galvanometer. Because the shunt resistance is deliberately made very small compared to the galvanometer resistance, the combined (parallel) resistance of the ammeter ends up being very close to the shunt resistance itself — a large resistance in parallel with a tiny one is dominated by the tiny one.

    Step-by-Step Solution

    1. Parallel resistance formula: $R = \dfrac{G\, r_s}{G + r_s}$. …

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