Two resistances of 400 Omega and 800 Omega | vedclass

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(D) Before connecting the voltmeter,the potential difference across the $400 \ \Omega$ resistance is calculated using the voltage divider rule:$V_i =

Vedclass · Accepted Answer

$0.05$

Vedclass · Answer

(D) Before connecting the voltmeter,the potential difference across the $400 \ \Omega$ resistance is calculated using the voltage divider rule:$V_i = \frac{400}{400 + 800} 	imes 6 = \frac{400}{1200} 	imes 6 = 2 \ V$After connecting the voltmeter,the $400 \ \Omega$ resistor and the $10,000 \ \Omega$ voltmeter are in parallel. The equivalent resistance $R_p$ between points $A$ and $B$ is:$R_p = \frac{400 	imes 10,000}{400 + 10,000} = \frac{4,000,000}{10,400} \approx 384.6 \ \Omega$The potential difference measured by the voltmeter $(V_f)$ is the voltage across this parallel combination:$V_f = \frac{R_p}{R_p + 800} 	imes 6 = \frac{384.6}{384.6 + 800} 	imes 6 = \frac{384.6}{1184.6} 	imes 6 \approx 1.948 \ V \approx 1.95 \ V$The error in measurement is the difference between the ideal and measured values:$	ext{Error} = V_i - V_f = 2 - 1.95 = 0.05 \ V$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

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