A moving coil meter has a resistance of 100 Ω and at 5 V it gives full scale deflection. Find the value of external resistance to be connected in series for measuring 300 V.
5900 Ω
A moving coil meter is a sensitive instrument that measures electric current. To convert it into a voltmeter, a high resistance needs to be connected in series with the meter. This external series resistance helps to extend the voltage range that the meter can measure. The problem asks us to find the value of this external resistance required to measure a higher voltage.
A moving coil meter (also known as a galvanometer) has a certain internal resistance and gives a full-scale deflection at a specific voltage or current. When converting this meter into a voltmeter, the main goal is to limit the current flowing through the meter coil when a higher voltage is applied, thus preventing damage and allowing it to measure the extended voltage range accurately. This is achieved by adding an external resistance in series.
First, we need to determine the current flowing through the moving coil meter when it shows full-scale deflection. This current is crucial because it's the maximum current the meter can safely handle.
Given:
Using Ohm's Law, the full-scale deflection current ($\text{I}_\text{g}$) can be calculated as:
$\text{I}_\text{g} = \frac{\text{V}_\text{m}}{\text{R}_\text{m}}$
$\text{I}_\text{g} = \frac{5 \, \text{V}}{100 \, \Omega}$
$\text{I}_\text{g} = 0.05 \, \text{A}$
This means that when the meter measures 5 V, a current of 0.05 A flows through it, causing full deflection. This is the maximum current it can handle for its specified range.
Now, we want the same moving coil meter to measure a maximum voltage of 300 V. For this extended range, the total resistance of the voltmeter (meter resistance + external series resistance) must be such that the full-scale deflection current ($0.05 \, \text{A}$) still flows through the meter when 300 V is applied across the entire combination.
Let the desired maximum voltage be $\text{V} = 300 \, \text{V}$.
The total resistance ($\text{R}_\text{total}$) required for this new range is:
$\text{R}_\text{total} = \frac{\text{V}}{\text{I}_\text{g}}$
$\text{R}_\text{total} = \frac{300 \, \text{V}}{0.05 \, \text{A}}$
$\text{R}_\text{total} = 6000 \, \Omega$
The total resistance ($\text{R}_\text{total}$) is the sum of the meter's internal resistance ($\text{R}_\text{m}$) and the external resistance ($\text{R}_\text{s}$) connected in series.
$\text{R}_\text{total} = \text{R}_\text{m} + \text{R}_\text{s}$
We need to find the value of the external resistance ($\text{R}_\text{s}$).
$\text{R}_\text{s} = \text{R}_\text{total} - \text{R}_\text{m}$
Substituting the calculated values:
$\text{R}_\text{s} = 6000 \, \Omega - 100 \, \Omega$
$\text{R}_\text{s} = 5900 \, \Omega$
Therefore, an external resistance of $5900 \, \Omega$ must be connected in series with the moving coil meter to enable it to measure up to 300 V. This ensures that when 300 V is applied, the current through the sensitive moving coil meter remains at its full-scale deflection value, 0.05 A, which it can safely handle.
| Parameter | Formula/Value | Calculation |
|---|---|---|
| Meter Resistance ($\text{R}_\text{m}$) | Given | $100 \, \Omega$ |
| Full Scale Deflection Voltage ($\text{V}_\text{m}$) | Given | $5 \, \text{V}$ |
| Target Voltage ($\text{V}$) | Given | $300 \, \text{V}$ |
| Full Scale Deflection Current ($\text{I}_\text{g}$) | $\frac{\text{V}_\text{m}}{\text{R}_\text{m}}$ | $\frac{5 \, \text{V}}{100 \, \Omega} = 0.05 \, \text{A}$ |
| Total Resistance for 300 V ($\text{R}_\text{total}$) | $\frac{\text{V}}{\text{I}_\text{g}}$ | $\frac{300 \, \text{V}}{0.05 \, \text{A}} = 6000 \, \Omega$ |
| External Series Resistance ($\text{R}_\text{s}$) | $\text{R}_\text{total} - \text{R}_\text{m}$ | $6000 \, \Omega - 100 \, \Omega = 5900 \, \Omega$ |
The calculated value of the external resistance is $5900 \, \Omega$.
If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter :
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Which of the following is correct for ammeter?
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