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(A) The ideal voltage across the $4 	ext{ k}\Omega$ resistor (without the voltmeter) is calculated using the voltage divider rule: $V = \frac{4 	ext

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$14$

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(A) The ideal voltage across the $4 	ext{ k}\Omega$ resistor (without the voltmeter) is calculated using the voltage divider rule: $V = \frac{4 	ext{ k}\Omega}{4 	ext{ k}\Omega + 4 	ext{ k}\Omega} 	imes 4 	ext{ V} = 2 	ext{ V}$.When the voltmeter with resistance $R_v = 12 	ext{ k}\Omega$ is connected in parallel with the $4 	ext{ k}\Omega$ resistor, the equivalent resistance $R_p$ of the parallel combination is: $R_p = \frac{4 	ext{ k}\Omega 	imes 12 	ext{ k}\Omega}{4 	ext{ k}\Omega + 12 	ext{ k}\Omega} = \frac{48}{16} 	ext{ k}\Omega = 3 	ext{ k}\Omega$.The actual reading $V^{\prime}$ of the voltmeter is the voltage across this parallel combination: $V^{\prime} = \frac{R_p}{R_p + 4 	ext{ k}\Omega} 	imes 4 	ext{ V} = \frac{3}{3 + 4} 	imes 4 	ext{ V} = \frac{12}{7} 	ext{ V} \approx 1.714 	ext{ V}$.The percentage error is given by: $	ext{Percentage Error} = \frac{V - V^{\prime}}{V} 	imes 100 = \frac{2 - \frac{12}{7}}{2} 	imes 100 = \frac{\frac{14 - 12}{7}}{2} 	imes 100 = \frac{2}{14} 	imes 100 = \frac{1}{7} 	imes 100 \approx 14.28 \%$.Thus, the percentage error is nearly $14 \%$.

The percentage error in the reading of the voltmeter in the figure shown here is nearly ............ $\%$

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