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Q.A voltmeter of resistance 1000 Ω1000\ \Omega can measure up to 25 V. How will you convert it so that it can read up to 250 V?

SIKKIM-CBSECBSE Class XII Board 2025Subjective· 2mImportance★★★★★
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To extend the range of a voltmeter from 25 V to 250 V, you must connect a high resistance in series with the existing meter. The required series resistance is 9000 Ω9000\ \Omega.


Why this works: The core idea

A voltmeter is essentially a sensitive current-measuring device (a galvanometer) with a large fixed resistor in series. The pointer deflects fully when a specific current — the full-scale deflection current — flows through it. For the given voltmeter, that full-scale current is:

Ig=Voltage rangeTotal resistance=25 V1000 Ω=0.025 A=25 mAI_g = \frac{\text{Voltage range}}{\text{Total resistance}} = \frac{25\ \text{V}}{1000\ \Omega} = 0.025\ \text{A} = 25\ \text{mA}

This current must never exceed 25 mA25\ \text{mA}, or the meter will be damaged. To measure a larger voltage (250 V), you need to limit the current to the same 25 mA25\ \text{mA} even when the applied voltage is ten times higher. The only way to do that is to add extra resistance in series — the current will then be I=VRtotalI = \frac{V}{R_{\text{total}}}, so a larger RtotalR_{\text{total}} keeps II at the safe value.

Important

The galvanometer’s full-scale deflection current IgI_g is the invariant in any voltmeter conversion. You never change the meter movement itself — you only add series resistance to drop the extra voltage.


Step-by-step solution

1. Find the full-scale deflection current of the existing voltmeter

The voltmeter has resistance Rv=1000 ΩR_v = 1000\ \Omega and measures up to Vv=25 VV_v = 25\ \text{V}. The current that produces full deflection is:

Ig=VvRv=251000=0.025 AI_g = \frac{V_v}{R_v} = \frac{25}{1000} = 0.025\ \text{A}

This IgI_g is the maximum current the meter can safely handle.

2. Determine the total resistance needed for the new range

You want the voltmeter to read up to Vnew=250 VV_{\text{new}} = 250\ \text{V}. For the same Ig=0.025 AI_g = 0.025\ \text{A}, the total series resistance required is:

Rtotal=VnewIg=2500.025=10,000 ΩR_{\text{total}} = \frac{V_{\text{new}}}{I_g} = \frac{250}{0.025} = 10{,}000\ \Omega

3. Calculate the extra series resistance to add

The existing voltmeter already contributes 1000 Ω1000\ \Omega. So the additional resistance RsR_s you must connect in series is:

Rs=Rtotal−Rv=10,000−1000=9000 ΩR_s = R_{\text{total}} - R_v = 10{,}000 - 1000 = 9000\ \Omega …

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