Question

A galvanometer having a resistance of 80 ohm is shunted by a wire of resistance 2 ohm. If the total current is two ampere, the part of it passing through the shunt is

• A. 0.25amp
• B. 1.95 amp
• C. 0.2amp
• D. 0.5amp

Answer 1

Answer: (B)

1.95 amp

$Req=\frac{80\times 2}{80+2}$

$\frac{i}{i_5}=\frac{R_s}{Req}=\frac{2}{i_s}=\frac{2\times 41}{80}$ (we have used the standard euation in case of galvanometer according to question)

$i_s=\frac{80}{41}=1.95A$