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A galvanometer with a 1 mA full scale deflection and an internal resistance of 500 ohm is to be used as voltmeter, find series resistance for 1 and 10 v ranges.
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Answer: Given : Full scale deflection current $(I_g) = 1 \times 10^{-3}A$

internal resistance $R_g= 500 \Omega$

Voltage range $V=1 V , \ V=18 V$

To find : series resistance - ?

Formula : $R=\dfrac{V}{I_g}-R_g$

Solution :  $R=\dfrac{V}{I_g}-R_g=\dfrac{1}{10^{-3}}-500=500\Omega$

$R=\dfrac{V}{I_g}-R_g=\dfrac{18}{10^{-3}}-500=17,500\Omega$

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