Question

A voltmeter of resistance 1000 $\Omega$ can measure up to 25 V. How will you convert it so that it can read up to 250 V ?

Answer 1

Connect a resistance of 9000 $\Omega$ in series with the voltmeter.

Answer 2

To convert a voltmeter to a higher range, a suitable resistance is connected in series with it.

1. Calculate the full-scale deflection current ($I$) of the voltmeter: The current flowing through the voltmeter when it reads its maximum voltage is given by Ohm's law: $I = \frac{V_{\text{meter}}}{R_{\text{meter}}}$ Given $V_{\text{meter}} = 25 \text{ V}$ and $R_{\text{meter}} = 1000 \text{ } \Omega$. $I = \frac{25 \text{ V}}{1000 \text{ } \Omega} = 0.025 \text{ A}$

2. Calculate the required series resistance ($R_s$): To measure a higher voltage $V_{\text{new}} = 250 \text{ V}$, the same full-scale current $I$ must flow through the voltmeter and the series resistance. The total resistance of the circuit will be $R_{\text{total}} = R_{\text{meter}} + R_s$. Using Ohm's law for the new range: $V_{\text{new}} = I (R_{\text{meter}} + R_s)$ $250 \text{ V} = 0.025 \text{ A} (1000 \text{ } \Omega + R_s)$

Rearranging the equation to solve for $R_s$: $1000 \text{ } \Omega + R_s = \frac{250 \text{ V}}{0.025 \text{ A}}$ $1000 \text{ } \Omega + R_s = 10000 \text{ } \Omega$ $R_s = 10000 \text{ } \Omega - 1000 \text{ } \Omega$ $R_s = 9000 \text{ } \Omega$

Alternatively, the voltage drop across the series resistor must be the difference between the new range and the old range: $V_s = V_{\text{new}} - V_{\text{meter}} = 250 \text{ V} - 25 \text{ V} = 225 \text{ V}$ The series resistance is then: $R_s = \frac{V_s}{I} = \frac{225 \text{ V}}{0.025 \text{ A}} = 9000 \text{ } \Omega$

Therefore, to convert the voltmeter to read up to 250 V, a resistance of 9000 $\Omega$ must be connected in series with it.