Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A galvanometer has a coil of resistance $100\Omega$ and gives a full scale deflection for $30 \text{ mA}$ current. If it is to work as a voltmeter of $30\text{ V}$ range, the resistance required to be added will be

900 $\Omega$

1800 $\Omega$

500 $\Omega$

1000 $\Omega$

Step-by-Step Solution

Concept: To convert a galvanometer into a voltmeter, a high resistance ( $R$ ) is connected in series with the galvanometer coil. This ensures that the total resistance is high enough to draw negligible current from the circuit being measured.
Formula: According to Ohm's Law, the voltage drop across the series combination must equal the desired range ( $V$ ) when the full-scale deflection current ( $I_g$ ) flows through it. The equation is: $V = I_g (R_G + R)$ Where:

$V$ is the voltage range ( $30 \text{ V}$ ).
$I_g$ is the full-scale deflection current ( $30 \text{ mA} = 30 \times 10^{-3} \text{ A}$ ).
$R_G$ is the galvanometer resistance ( $100 \Omega$ ).
$R$ is the external resistance to be added .

Calculation: Rearranging the formula for $R$ : $R = \frac{V}{I_g} - R_G$ $R = \frac{30}{30 \times 10^{-3}} - 100$ $R = \frac{1}{10^{-3}} - 100$ $R = 1000 - 100 = 900 \Omega$

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