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    NEET 2026 Preparation Tips for Chemistry, Biology and Physics

    Voltmeter MCQ - Practice Questions with Answers

    Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

    Quick Facts

    • Voltmeter is considered one of the most asked concept.

    • 18 Questions around this concept.

    Solve by difficulty

    A galvanometer of resistance $100 \Omega$ has 50 divisions on its scale and has sensitivity of $20 \mu A /$ division. It is to be converted to a voltmeter with three ranges, of $0-2 \mathrm{~V}, 0-10 \mathrm{~V}$ and $0-20 \mathrm{~V}$. The appropriate circuit to do so is :

     A galvanometer , whose resistance is 50 ohm , has 25 divisions in it. When a current of 4 X 10-4 A passes through it , its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V , it should be connected to a resistance (in ohm)  of : 

    The actual value of resistance $R$, shown in the figure is $30 \Omega$. This is measured in an experiment as shown using the standard formula $R=\frac{V}{I}$, Where } V and } I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is $5 \%$ less, then the internal resistance (in $\Omega$ ) of the voltmeter is: (Round the answer to the nearest integer)

    An ideal voltmeter V is connected to a $20 \Omega$ resistor and a battery with emf 10V and internal resistance $0.5 \Omega$ as in the figure.

    The ratio of current in $20 \Omega$ resistor with the voltage of the battery

    In the given circuit, the voltmeter and the electric cell are ideal. Find the reading of the voltmeter

    A constant 60 V d.c. supply is connected across two resistors of resistance $400 \mathrm{k} \Omega$ and $200 \mathrm{k} \Omega$. What is the reading of the voltmeter, also of resistance $200 \mathrm{k} \Omega$, when connected across the second resistor as shown in Figure?

    An ammeter A of finite resistance and a resistor R are joined in series to an ideal cell C. A Potentiometer P is joined in parallel to R. The ammeter reading is $\mathrm{I}_0$ and the potentiometer reading is $\mathrm{V}_0$. P is now replaced by a voltmeter of finite resistance. The ammeter reading now is I and the voltmeter reading is V. Then

     

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    A constant voltage dc source is connected, as shown in Figure, across two resistors of resistances $400 \mathrm{k} \Omega$ and $100 \mathrm{k} \Omega$

    What is the reading of the voltmeter, also of resistance $100 \mathrm{k} \Omega$, when connected across the second resistor as shown?

    A, B and C are voltmeters of resistance R, 1.5 R and 3 R, respectively. When some potential difference is applied between X and Y, the voltmeter reading are \mathrm{V_A, V_B} and \mathrm{V_C}, respectively. Then

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    A voltmeter having a resistance of $1800 \Omega$ is employed to measure the potential difference across $200 \Omega$ resistance which is connected to be power supply of 50 V and internal resistance $20 \Omega$. What is the approximate percentage change in the p.d. across $200 \Omega$ resistance as a result of connecting the voltmeter across it?

    Concepts Covered - 1

    Voltmeter

    Voltmeter

    It is a device used to measure Potential difference and is always put in parallel with the circuit element whose voltage is to be measured

    Conversion of galvanometer into voltmeter:  by Connecting a large Resistance R  in series. 

    - The equivalent resistance of the combination is $G+R$
    - The required value of high resistance to be connected in series with the galvanometer is

    $
    R=\frac{V}{I_g}-G=\left[\frac{V}{V_g}-1\right] G_{. \mathrm{V} \text { is the total voltage applied across the circuit and }} V_g \text { is the total }
    $
     voltage drop across the galvanometer.
    - if $n^{\text {th }}$ part of the applied voltage across galvanometer, that is $V_g=V / n$, then $R=(n-1) G$.

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    Voltmeter

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