The ratio of heats generated through shunt an | vedclass

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Question

(A) Given,the ratio of heats generated through the shunt $(H_s)$ and galvanometer $(H_g)$ is $H_s : H_g = 7 : 5$.The resistance of the galvanometer is

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) Given,the ratio of heats generated through the shunt $(H_s)$ and galvanometer $(H_g)$ is $H_s : H_g = 7 : 5$.The resistance of the galvanometer is $R_g = 112 \Omega$.Since the shunt and galvanometer are connected in parallel to form an ammeter,the potential difference $(V)$ across both is the same.The heat generated in a resistor is given by $H = \frac{V^2}{R} 	imes t$.Therefore,the ratio of heat generated is $\frac{H_s}{H_g} = \frac{V^2 / R_s}{V^2 / R_g} = \frac{R_g}{R_s}$.Given $\frac{H_s}{H_g} = \frac{7}{5}$,we have $\frac{R_g}{R_s} = \frac{7}{5}$.Substituting the value of $R_g = 112 \Omega$:$\frac{112}{R_s} = \frac{7}{5}$$R_s = \frac{112 	imes 5}{7}$$R_s = 16 	imes 5 = 80 \Omega$.Thus,the resistance of the shunt is $80 \Omega$.

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: (in $\Omega$)

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