When a voltmeter with finite resistance ($R_v = 900 \, \Omega$) is connected in parallel with a resistor ($R = 100 \, \Omega$), the equivalent resistance of the combination decreases to $R_{eq} = \frac{R R_v}{R + R_v} = \frac{100 \times 900}{100 + 900} = 90 \, \Omega$ . This reduction in resistance draws more current (loading effect) or alters the potential division in the circuit compared to the ideal case where the voltmeter has infinite resistance. The percentage error is calculated as $\frac{V_{actual} - V_{measured}}{V_{actual}} \times 100$. For an error of $1.0\%$, the circuit typically implies a series resistance (e.g., $10 \, \Omega$) such that the potential drop changes from the ideal value by this percentage .