The ratio of heats generated through shunt and galvanometer is \(7: 5\) when they are connected to make an…
The ratio of heats generated through shunt and galvanometer is \(7: 5\) when they are connected to make an ammeter. If the resistance of the galvanometer is \(112 \Omega\) then the resistance of the shunt is
\(80 \Omega\)
\(8 \Omega\)
\(15.6 \Omega\)
\(1.56 \Omega\)
Solution
Given, the ratio of heats generated through the shunt and galvanometer is \(7: 5\), and resistance of galvanometer, \(R_g=112 \Omega\)
\(\therefore \frac{\text { Heat generated through shunt }\left(p_s\right)}{\text { Heat generated through galvanometer }\left(p_g\right)}=\frac{7}{5}\)
\(\begin{aligned}
& \therefore \quad \frac{\frac{V^2}{R_s}}{\frac{V^2}{R_g}}=\frac{7}{5} \quad\left(\because p_S=\frac{V^2}{R_S} \text { and } p_g=\frac{V^2}{R_g}\right) \\
& R_s=\frac{5}{7} \times R_g \Rightarrow R_s=\frac{5}{7} \times 112=80 \Omega \\
\end{aligned}\)
Hence, the resistance of shunt, \(R_s=80 \Omega\)