$A$ galvanometer can be converted to a voltmeter of full-scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full-scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$. What is the current flowing through the galvanometer at its full-scale deflection?

  • A
    $\frac{V_{0}-I_{0} R_{2}}{R_{1}-R_{2}}$
  • B
    $\frac{V_{0}+I_{0} R_{2}}{R_{1}+R_{2}}$
  • C
    $\frac{V_{0}-I_{0} R_{2}}{R_{2}-R_{1}}$
  • D
    $\frac{V_{0}+I_{0} R_{1}}{R_{1}+R_{2}}$

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$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)

When a resistance of $5\,\Omega$ is shunted with a moving coil galvanometer,it shows a full scale deflection for a current of $250\,mA$. However,when a $1050\,\Omega$ resistance is connected with it in series,it gives full scale deflection for $25\,V$. The resistance of the galvanometer is $......\,\Omega$.

$A$ galvanometer gives full scale deflection with $0.006 \ A$ current. By connecting it to a $4990 \ \Omega$ resistance,it can be converted into a voltmeter of range $0-30 \ V$. If connected to a $\frac{2n}{249} \ \Omega$ resistance,it becomes an ammeter of range $0-1.5 \ A$. The value of $n$ is:

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