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Question
A voltmeter coil has resistance 50.0 Ω and a resistor of 1.15 kΩ is connected in series. It can read potential differences up to 12 volts. If this same coil is used to construct an ammeter that can measure currents up to 2.0 A, what should be the resistance of the shunt used?
Solution
The effective resistance for the voltmeter,
Reff = (1150 + 50) Ω = 1200 Ω
The maximum current ig through the coil for maximum deflection of 12 V,
\[i_g = \left( \frac{12}{1200} \right) A = 0 . 01 A\]
Let R be the resistance of the shunt used for ammeter.
Resistance of the coil, g = 50 Ω
Maximum deflection current, i, through the ammeter = 2 A
Since R and g are connected in parallel,
\[g \times i_g = R \times \left( i - i_g \right)\]
\[ \Rightarrow R = \frac{g \times i_g}{\left( i - i_g \right)}\]
\[ \Rightarrow R = \frac{50 \times 0 . 01}{2 - 0 . 01} = 0 . 251 \Omega\]
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