Question

Circuit shown in figure is a simple ohm – meter, in which G is a galvanometer (of very small resistance compared to R and $R_o$ ), $R_o$ is a known resistance and R is the resistance which is to be measured. If A and B are short circuited by a resistance less wire, galvanometer gives full scale deflection. Then to read the resistance R directly from galvanometer, its scale would look like

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

. Suppose e.m.f of the battery connected in the circuit is equal to E. Then the current at R = O will be equal to $I_0=\frac{E}{R_0}$ , when R increases, current through the circuit decreases.
If $R=R_0$, then net resistance of the circuit will become equal to $2R_0$. Hence current through the circuit will become $\frac{E}{2R_0}=\frac{I_0}{2}$.

To reduce the current to $\frac{I_0}{4}$, the net resistance of the circuit is to be increased to 4 times of its original value $R_0$. Hence $R=3R_0$.

Similarly, to reduce the current to $\frac{I_0}{8}$, the resistance R must be equal to $7R_0$ and to reduce the current to zero, R must be very large. Hence A is correct.