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(A) Case $1$: When the resistance of the voltmeter is considered infinite (ideal voltmeter).In this case,the voltmeter does not draw any current. The

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$16.67$

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(A) Case $1$: When the resistance of the voltmeter is considered infinite (ideal voltmeter).In this case,the voltmeter does not draw any current. The circuit consists of a $600 \, \Omega$ resistor and a $400 \, \Omega$ resistor in series connected to a $12 \, V$ source.The voltage across the $400 \, \Omega$ resistor is given by the voltage divider rule:$V_{1} = 12 	imes \frac{400}{600 + 400} = 12 	imes \frac{400}{1000} = 4.8 \, V$.Case $2$: When the resistance of the voltmeter $(1200 \, \Omega)$ is taken into account.The voltmeter is in parallel with the $400 \, \Omega$ resistor. The equivalent resistance of this parallel combination is:$R_{p} = \frac{400 	imes 1200}{400 + 1200} = \frac{480000}{1600} = 300 \, \Omega$.Now,the circuit consists of a $600 \, \Omega$ resistor in series with the $300 \, \Omega$ equivalent resistance.The voltage across the parallel combination (which is the voltmeter reading) is:$V_{2} = 12 	imes \frac{300}{600 + 300} = 12 	imes \frac{300}{900} = 4 \, V$.The percentage error is calculated as:$	ext{Error} = \frac{V_{1} - V_{2}}{V_{1}} 	imes 100 \% = \frac{4.8 - 4}{4.8} 	imes 100 \% = \frac{0.8}{4.8} 	imes 100 \% = \frac{1}{6} 	imes 100 \% = 16.67 \%$.

Find the percentage error in the reading of the voltmeter shown in the circuit. (in $\%$)

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