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(B) In circuit $(a)$,the voltmeter is in series with the $1 \,\Omega$ resistor,and the ammeter is in parallel with the $1 \,\Omega$ resistor. Let $R_V

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$10^3$ and $10^{-3}$

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(B) In circuit $(a)$,the voltmeter is in series with the $1 \,\Omega$ resistor,and the ammeter is in parallel with the $1 \,\Omega$ resistor. Let $R_V$ be the resistance of the voltmeter and $R_A$ be the resistance of the ammeter. The voltage across the $1 \,\Omega$ resistor is $V = I_1 	imes 1$. The current through the ammeter is $I_2 = V / R_A = I_1 / R_A$. The total current $I = I_1 + I_2 = I_1(1 + 1/R_A)$. The voltmeter reading is $V_m = I 	imes R_V$. The ratio of voltage to current is $V_m / I = R_V = 1000 \,\Omega$ (assuming the voltmeter measures the total voltage drop across the series combination). However,based on the circuit,the ratio $V/I$ given is $1000 \,\Omega$. From the circuit $(a)$,the effective resistance is $R_{eq} = R_V + (1 \parallel R_A) \approx R_V = 1000 \,\Omega$.In circuit $(b)$,the ammeter is in series with the $1 \,\Omega$ resistor,and the voltmeter is in parallel with the $1 \,\Omega$ resistor. The effective resistance is $R_{eq} = R_A + (1 \parallel R_V) \approx R_A + 1 = 0.999 \,\Omega$. Since $R_V$ is large,$1 \parallel R_V \approx 1 \,\Omega$. Thus $R_A + 1 = 0.999$,which implies $R_A \approx 0$ (or $10^{-3} \,\Omega$).Thus,$R_V = 10^3 \,\Omega$ and $R_A = 10^{-3} \,\Omega$.

Similar Questions

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