Careers360 Logo
ask-icon
share
    Re NEET 2026 Marks vs Rank Expected: AIR, Safe Score, Percentile

    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

     

    JSS University Mysore Allied Sciences 2026

    NAAC A+ Accredited| Ranked #21 in University Category by NIRF | Applications open for multiple UG & PG Programs

    Emversity Allied Health Programs

    Get Job Ready in Healthcare | Employability-Focused Programs

    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

    NEET 2026 Rank Predictor
    Predict your NEET 2026 rank instantly! Enter your expected score and get an estimated AIR, percentile, and college admission chances with NEET 2026 Rank Predictor.
    Try Now

    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 $R=(n-1) G$.

    Study it with Videos

    Voltmeter

    "Stay in the loop. Receive exam news, study resources, and expert advice!"

    Get Answer to all your questions