Two resistances of 400 Omega and 800 Omega ar | vedclass

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(D) $1$. Potential difference $(V_1)$ across the $400 \Omega$ resistor without the voltmeter:$V_1 = \frac{R_1}{R_1 + R_2} 	imes V = \frac{400}{400 +

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$0.05$

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(D) $1$. Potential difference $(V_1)$ across the $400 \Omega$ resistor without the voltmeter:$V_1 = \frac{R_1}{R_1 + R_2} 	imes V = \frac{400}{400 + 800} 	imes 6 = \frac{400}{1200} 	imes 6 = 2 	ext{ V}$.$2$. When the voltmeter of resistance $R_v = 10000 \Omega$ is connected in parallel with the $400 \Omega$ resistor,the equivalent resistance $(R_p)$ of this parallel combination is:$R_p = \frac{400 	imes 10000}{400 + 10000} = \frac{4000000}{10400} = \frac{40000}{104} \approx 384.62 \Omega$.$3$. The new potential difference $(V_2)$ across the parallel combination is:$V_2 = \frac{R_p}{R_p + R_2} 	imes V = \frac{384.62}{384.62 + 800} 	imes 6 = \frac{384.62}{1184.62} 	imes 6 \approx 1.948 	ext{ V}$.$4$. The error in measurement is:$	ext{Error} = V_1 - V_2 = 2 - 1.948 = 0.052 	ext{ V}$.Rounding to the nearest given option,the error is approximately $0.05 	ext{ V}$.

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