Question

A galvanometer has a current sensitivity of $1mA$ per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of $500\Omega$ and cell of internal resistance $1\Omega$. It gives a deflection of 5 division for shunt of $5$ ohm and $20$ division for shunt of $25$ ohm. The emf of cell is

• A. 47.1 V
• B. 57.1 V
• C. 67.1 V
• D. 77.1 V

Answer 1

Answer: (A)

Here, $I=\frac{E}{R+r+\frac{GS}{G+S}}$ and $I_g=\frac{IS}{G+S}$
$I_g=\frac{E}{\left(R+r\right)+\frac{GS}{\left(G+S\right)}}\times \frac{S}{\left(G+S\right)}$
$\therefore I_g=\frac{ES}{\left(R+r\right)\left(G+S\right)+GS}$
For $S=5$ohm, $I_g=5\times 10^{-3}A$ and for $S=25$ ohm,
$I_g=20\times 10^{-3}$A
Hence, $5\times 10^{-3}=\frac{E\times 5}{501\left(G+25\right)+5G}...\left(i\right)$
and $20\times 10^{-3}=\frac{E\times 5}{501\left(G+5\right)+25G}...\left(ii\right)$
Dividing and solving,
$G=88.2\Omega$
From $\left(i\right),$we get
$E=10^{-3}\left[501\left(88.2+5\right)+5\times 88.2\right]$
$=47.1volt$