A uniform resistance wire of length L and total resistance 100 is connected across an ideal 10 V battery. A voltmeter with a finite internal resistance of 100 is used to measure the potential difference across the first half of the wire (length L/2 ). What is the percentage error in the voltmeter reading compared to the true potential difference that would exist without the voltmeter connected?

Step-by-step solution

Without the voltmeter, the wire acts as two 50 resistors in series. The true potential difference across the first half is V_ true = 10 50 100 = 5 V . When the voltmeter is connected, it is in parallel with the first 50 segment. The equivalent resistance of this parallel combination is R_p = 100 50 100 + 50 = 100 3 . The total resistance of the circuit becomes R_ total = R_p + 50 = 100 3 + 50 = 250 3 . The voltage measured by the voltmeter is V_ measured = 10 R_p R_ total = 10 100/3 250/3 = 4 V . The absolute error