A voltmeter having an internal resistance of 900\ \Omega is connected across the terminals… — JAMB UTME

A voltmeter having an internal resistance of $900\ \Omega$ is connected across the terminals of a cell with an electromotive force (e.m.f.) of $1.50\text{ V}$ and an internal resistance of $100\ \Omega$ . What is the reading on the voltmeter?

$1.35\text{ V}$ Answer
B
$1.50\text{ V}$
C
$0.15\text{ V}$
D
$0.17\text{ V}$

Answer

The reading on the voltmeter is

1.35\text{ V}

When connected across the cell, the voltmeter's resistance forms a series circuit with the cell's internal resistance. The total resistance is

900\ \Omega + 100\ \Omega = 1000\ \Omega

. The current drawn from the cell is

I = \frac{1.50\text{ V}}{1000\ \Omega} = 0.0015\text{ A}

. The voltage indicated by the meter is the potential difference across its terminals,

V = 0.0015\text{ A} \times 900\ \Omega = 1.35\text{ V}

(or

E - Ir = 1.50 - 0.15 = 1.35\text{ V}

Step-by-Step Solution

Calculate the total resistance of the circuit.

R_{\text{total}} = R_v + r = 900\ \Omega + 100\ \Omega = 1000\ \Omega

The voltmeter resistance and the internal resistance of the cell are connected in series.

Calculate the current flowing in the circuit using Ohm's law.

I = \frac{E}{R_{\text{total}}} = \frac{1.50\text{ V}}{1000\ \Omega} = 0.0015\text{ A}

The electromotive force drives current through the total circuit resistance.

Calculate the potential difference across the voltmeter (terminal potential difference).

V = I \times R_v = 0.0015\text{ A} \times 900\ \Omega = 1.35\text{ V}

The voltmeter measures the potential drop across its own internal resistance.

Key Concept

Terminal Potential Difference and Voltmeter Loading Effect

Estimated Time:1m 30s

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