Question

If there is no deflection in the galvanometer connected in a circuit shown in figure, then the ratio of lengths AC/CB is

• A. 4 : 1
• B. 1 : 4
• C. 1:1
• D. 2 : 1

Answer 1

Answer: (A)

Let $AC=l_1,CB=l_2$ and r the resistance per unit length of wire AB. As there is no current in galvanometer G, the resistance 60 Q, 15 0, AC and CB form the arms of a balanced Wheatstone's bridge $\therefore$ $\frac{P}{Q}=\frac{R}{S}$ $\Rightarrow$ $\frac{60}{15}=\frac{r\times l_1}{r\times l_2}$ Or $\frac{l_1}{l_2}=\frac{4}{1}$