Two resistances of 400,Omega and 800,Omega ar | vedclass

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Question

(D) The actual voltage drop across the $400\,\Omega$ resistor without the voltmeter is calculated using the voltage divider rule:$V_{actual} = V 	ime

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$0.05$

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(D) The actual voltage drop across the $400\,\Omega$ resistor without the voltmeter is calculated using the voltage divider rule:$V_{actual} = V 	imes \frac{R_1}{R_1 + R_2} = 6 	imes \frac{400}{400 + 800} = 6 	imes \frac{400}{1200} = 6 	imes \frac{1}{3} = 2\,V$.When a voltmeter of resistance $R_v = 10,000\,\Omega$ is connected in parallel with the $400\,\Omega$ resistor,the equivalent resistance $R_p$ of this parallel combination is:$R_p = \frac{400 	imes 10,000}{400 + 10,000} = \frac{4,000,000}{10,400} = \frac{10,000}{26} \approx 384.62\,\Omega$.The new total resistance of the circuit is $R_{total} = R_p + 800 = 384.62 + 800 = 1184.62\,\Omega$.The current in the circuit is $I = \frac{V}{R_{total}} = \frac{6}{1184.62} \approx 0.005065\,A$.The measured voltage $V_{measured}$ is the voltage drop across $R_p$:$V_{measured} = I 	imes R_p = 0.005065 	imes 384.62 \approx 1.9487\,V$.The error in measurement is $\Delta V = V_{actual} - V_{measured} = 2 - 1.9487 = 0.0513\,V$.Rounding to the nearest option,the error is approximately $0.05\,V$.

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