CCommons
Search✍️ WriteSign in
⚛️ NEET Physics❓ Question

Converting a galvanometer to ammeter/voltmeter: why is the ammeter low-resistance and why is the scale linear?

Kavya NaikDropper Sample

10 Sept 2026 · 1 min read

Amma says I should explain things to myself out loud, so here goes, and then you tell me where I'm wrong.

A moving coil galvanometer has resistance G=100 ΩG = 100\ \Omega and full-scale deflection current Ig=5I_g = 5 mA.

To make a 0–1 A ammeter I connect a shunt SS in parallel: S=IgGI−Ig=0.005×1000.995≈0.50 ΩS = \dfrac{I_g G}{I - I_g} = \dfrac{0.005\times100}{0.995}\approx0.50\ \Omega.

To make a 0–10 V voltmeter I connect a resistance RR in series: R=VIg−G=100.005−100=1900 ΩR = \dfrac{V}{I_g} - G = \dfrac{10}{0.005}-100 = 1900\ \Omega.

My doubts:

  1. The converted ammeter's resistance. Is it SS or something else? And why do we want low resistance for an ammeter and high for a voltmeter?
  2. NCERT says the radial magnetic field makes the deflection proportional to current. I understand the torque is NIABNIAB but where does the radial field come in and what would go wrong without it?
  3. Current sensitivity and voltage sensitivity. If I increase the number of turns NN to make the galvanometer more sensitive, does it improve both?
WhatsAppPost on X

7 answers & replies

Sign in to join the conversation.

  • [+]Siddharth RaoDropperSample· 1mo ago

    Your numbers are right. 1. Resistance of the converted ammeter is SS and GG in parallel: RA=GSG+S=100×0.5025100.5025≈0.50 ΩR_A = \frac{GS}{G+S} = \frac{100\times0.5025}{100.5025}\approx0.50\ \Omega which is slightly less than SS, effectively about SS. Ammeter goes in series in the circuit, so we want it to add as little resistance as possible, otherwise the current we are trying to measure changes. Voltmeter goes in parallel across the component, so it should draw almost no current from the circuit, i.e. it needs very high resistance. Here RV=G+R=2000 ΩR_V=G+R=2000\ \Omega for the 10 V range.

    • [+]Kavya NaikDropperSample· 1mo agoAuthor

      Everything is cleared, and thanks to Meera for the radial field explanation. Writing "ammeter low, voltmeter high" with the arrows for in series / in parallel on my wall chart.

  • [+]Tushar PawarDropperSample· 1mo ago

    For the shunt you connect it in series with the galvanometer so that it takes extra current.

    • [+]Aman YadavDropperSample· 1mo ago

      Wrong. Shunt is in parallel. It must provide a path for the extra current to bypass the coil. Series resistor is used for voltmeter. Easy: "ammeter, shunt, parallel. Voltmeter, series resistor."

      • [+]Tushar PawarDropperSample· 1mo ago

        Haha yes, I flipped them. Shunt parallel, high resistance in series. Ok.

  • [+]Meera Patel11thSample· 1mo ago

    Radial field: the pole pieces are curved and there is a soft iron cylinder in the middle, so the field lines are along the radius, always in the plane of the coil. Whatever angle the coil turns, the field stays parallel to the plane of the coil (the normal makes 90° with the field). So the torque is always τ=NIAB\tau = NIAB (not NIABsin⁡αNIAB\sin\alpha) and does not depend on the angle. Restoring torque from the spring is kϕk\phi. At equilibrium, NIAB=kϕ⇒ϕ=(NABk)INIAB = k\phi \Rightarrow \phi = \left(\frac{NAB}{k}\right)I So deflection is proportional to the current, and the scale is uniform. Without the radial field the torque would change with angle and the scale wouldn't be linear.

  • [+]Nisha Verma12thSample· 1mo ago

    Sensitivity: current sensitivity ϕI=NABk\dfrac{\phi}{I}=\dfrac{NAB}{k} increases with NN. Voltage sensitivity ϕV=NABkR\dfrac{\phi}{V}=\dfrac{NAB}{kR} where RR is the coil's resistance. More turns means more length of wire so RR rises too (roughly R∝NR\propto N) and the ratio N/RN/R stays about the same. So increasing the number of turns increases current sensitivity but need not increase voltage sensitivity. This is a classic NCERT point that appears in assertion-reason type questions.

More from Kavya Naik

Keep reading