Q173 / 180 · KCET 2022 · Physics

A galvanometer of resistance $50 \Omega$ is connected to a battery $3 \mathrm{~V}$ along with a resistance $2950 \Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

  • A
    $5550 \Omega$
  • B
    $5050 \Omega$
  • Correct answer
    C
    $4450 \Omega$
  • D
    $6050 \Omega$

Solution

Resistance of galvanometer, $R_g=50 \Omega$ Emf of battery, $V=3 \mathrm{~V}$ Resistance connected in series, $R_s=2950 \Omega$ Total resistance, $R^{\prime}=R_g+R_s=50+2950=3000 \Omega$ $\therefore$ Current, $I=\frac{V}{R^{\prime}}=\frac{3}{3000}=10^{-3} \mathrm{~A}$ If the deflection has to be reduced to 20 divisions, then current, $I^{\prime}=\frac{I}{30} \times 20=\frac{2}{3} \times 10^{-3} \mathrm{~A}$ Let $R_E$ be the effective resistance of the circuit, hence $$ \begin{aligned} 3 & =R_E I^{\prime} \\ \Rightarrow \quad R_E & =\frac{3}{I^{\prime}}=\frac{3}{\frac{2}{3} \times 10^{-3}}=4.5 \times 10^3=4500 \Omega \end{aligned} $$ $$ \begin{aligned} \therefore \text { Resistance to be added } & =R_E-R_g \\ & =4500-50=4450 \Omega \end{aligned} $$

Hence, the correct answer is option: 'C'.

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