Question

In an experiment to determine the resistance of a galvanometer by half deflection method, the circuit shown is used. In one set of readings, if R = 10$\Omega$ and S =4$\Omega$, then the resistance of the galvanometer is

• A. $\frac{20}{3}\Omega$
• B. $\frac{40}{3}\Omega$
• C. $\frac{50}{3}\Omega$
• D. $\frac{10}{3}\Omega$

Answer 1

The correct option is A $\frac{20}{3}\Omega$

When $K_1$ is closed, the current through the galvanometer is $I_G=\frac{E}{R+G}.......\left(1\right)$

If $\theta$ is the deflection shown in galvanometer due to flow of current $I_G$, then $I_G\propto \theta$ or $I_G=k\theta$ where k is known as figure of merit.

so, using (1) $\frac{E}{R+G}=k\theta .....\left(2\right)$

When $K_2$ is also closed, the current through galvanometer is $I_G^{\prime }=\frac{S}{G+S}I$ where $I=$ current flow in the main circuit.

Now the value of S is adjusted such that the deflection $\theta$ will be half i.e $\frac{\theta }{2}$

Thus, $I_G^{\prime }=k\frac{\theta }{2}$

or $\frac{S}{G+S}I=k\frac{\theta }{2}........\left(3\right)$

Total resistance of the circuit is $R_t=R+\frac{SG}{S+G}$

Thus, $I=\frac{E}{R_t}=\frac{E}{R+\frac{SG}{S+G}}....\left(4\right)$

$\left(3\right)/\left(2\right)\Rightarrow \frac{SI(R+G)}{(G+S)E}=\frac{1}{2}$

now substituting the value of I then after solving we get the resistance of galvanometer is $G=\frac{RS}{R-S}$

or $G=\frac{10\times 4}{10-4}=\frac{20}{3}\Omega$