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(B) Given: Battery voltage $E = 6\,V$,high resistance $R = 11\,k\Omega = 11000\,\Omega$,figure of merit $k = 60\,\mu A/	ext{div}$,initial deflection

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(B) Given: Battery voltage $E = 6\,V$,high resistance $R = 11\,k\Omega = 11000\,\Omega$,figure of merit $k = 60\,\mu A/	ext{div}$,initial deflection $	heta = 9\,	ext{div}$.$1$. Calculate the current $I$ for full deflection $	heta = 9$:$I = k \cdot 	heta = 60 	imes 10^{-6} 	imes 9 = 540 	imes 10^{-6} = 5.4 	imes 10^{-4}\,A$.$2$. Using Ohm's law for the circuit without shunt:$I = \frac{E}{R + G} \implies 5.4 	imes 10^{-4} = \frac{6}{11000 + G}$.Since $R \gg G$,$R + G \approx R = 11000\,\Omega$.$G = \frac{E}{I} - R = \frac{6}{5.4 	imes 10^{-4}} - 11000 = 11111 - 11000 = 111.1\,\Omega$.$3$. In the half-deflection method,the shunt resistance $S$ is connected such that the deflection becomes $	heta/2 = 4.5\,	ext{div}$.The formula for shunt resistance in the half-deflection method is $S = \frac{G \cdot R}{R - G}$.Given $R = 11000\,\Omega$ and $G \approx 111.1\,\Omega$:$S = \frac{111.1 	imes 11000}{11000 - 111.1} \approx \frac{111.1 	imes 11000}{10888.9} \approx 112.2\,\Omega$.Rounding to the nearest given option,the value is $110\,\Omega$.

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