Question

A galvanometer having a resistance of 8 ohms is shunted by a wire of resistance 2 ohms. If the total current is 1 amp., the part of it passing through the shunt will be :

• A. 0.2 amp
• B. 0.25 amp
• C. 0.8 amp
• D. 0.5 amp

Answer 1

Answer: (C)

0.8 amp

Here, $V_G=V_S$

so $I_{G}G=I_{S}S$

or $(I-I_S)G=I_{S}S$

or $I_S=\frac{IG}{G+S}=\frac{1\times 8}{8+2}=0.8A$