Question

A galvanometer of 24ohm resistance can carry a full load of 5 micro ampere.If it is shunted by a resistance of 3ohm.How much current can this system carry now without damage??

Answer 1

$$\begin{gathered} asweknow \\ {R}_{sh}=\frac{R_m}{{\displaystyle \frac{I}{I_m}}-1} \\ \frac{I}{I_m}-1=\frac{R_m}{R_{sh}} \\ I=(\frac{R_m}{R_{sh}}+1)I_m \\ I=(\frac{24}{3}+1)\times 5\mu ampere \\ I=45\mu ampereanswer... \end{gathered}$$