A 50,Omega resistance is connected to a batte | vedclass

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(D) The current in the circuit without the ammeter is $I = \frac{V}{R} = \frac{5\,V}{50\,\Omega} = 0.1\,A$.The measured current $I'$ must be within $1

Vedclass · Accepted Answer

$r_s = 0.5\,\Omega$ in parallel with the galvanometer

Vedclass · Answer

(D) The current in the circuit without the ammeter is $I = \frac{V}{R} = \frac{5\,V}{50\,\Omega} = 0.1\,A$.The measured current $I'$ must be within $1\%$ of $I$. Thus,$I' \geq 0.99 	imes 0.1 = 0.099\,A$.To measure current,an ammeter (galvanometer with a shunt resistor $r_s$ in parallel) is connected in series with the $50\,\Omega$ resistor.The equivalent resistance of the circuit is $R_{eq} = 50 + R_A$,where $R_A = \frac{100 	imes r_s}{100 + r_s}$.The current measured is $I' = \frac{V}{R_{eq}} = \frac{5}{50 + R_A} = 0.099\,A$.Solving for $R_A$: $50 + R_A = \frac{5}{0.099} \approx 50.505\,\Omega$.So,$R_A = 50.505 - 50 = 0.505\,\Omega$.Since $R_A = \frac{100 	imes r_s}{100 + r_s} = 0.505$,we get $100 r_s = 50.5 + 0.505 r_s$.$99.495 r_s = 50.5 \Rightarrow r_s \approx 0.507\,\Omega$.Rounding to the nearest provided option,$r_s = 0.5\,\Omega$ connected in parallel is the correct choice.

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