The reading of the voltmeter in the circuit s | vedclass

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(A) First,we observe that the $60 \, \Omega$ resistor and the $40 \, \Omega$ voltmeter are connected in parallel.The equivalent resistance $R_p$ of th

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$2$

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(A) First,we observe that the $60 \, \Omega$ resistor and the $40 \, \Omega$ voltmeter are connected in parallel.The equivalent resistance $R_p$ of this parallel combination is given by:$R_p = \frac{60 	imes 40}{60 + 40} = \frac{2400}{100} = 24 \, \Omega$.Now,the circuit consists of a $40 \, \Omega$ resistor in series with this equivalent resistance $R_p = 24 \, \Omega$ across a $6 \, 	ext{V}$ battery.The total resistance of the circuit is $R_{eq} = 40 \, \Omega + 24 \, \Omega = 64 \, \Omega$.The voltage across the parallel combination (which is the reading of the voltmeter) can be found using the voltage divider rule:$V = V_{total} 	imes \frac{R_p}{R_{total}} = 6 	imes \frac{24}{64} = 6 	imes \frac{3}{8} = \frac{18}{8} = 2.25 \, 	ext{V}$.

The reading of the voltmeter in the circuit shown is ............. $V$. (in $.25$)

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