A galvanometer with a coil resistance of $80 \Omega$ deflects full-scale for a current of $2.0 \mathrm {~mA}$. What series resistance is required to convert it to a voltmeter reading full scale for $200 \mathrm {~V}$.
A) $100 \mathrm {~m} \Omega$
B) $100 \mathrm { k } \Omega$
C) $13 \mathrm { M } \Omega$
D) $0.80 \mathrm {~m} \Omega$
E) $250 \mathrm { k } \Omega$

Question

Quizplus · Accepted Answer

To convert a galvanometer to a voltmeter that reads full scale for 200 V with a full-scale deflection current of 2.0 mA and a coil resistance of 80 Ω, the series resistance needed is calculated using Ohm's law: $V = I \times R$. The total resistance required for 200 V is $R_{total} = \frac{V}{I} = \frac{200}{2.0 \times 10^{-3}} = 100,000 \Omega$ or $100 k\Omega$. Subtracting the internal resistance of the galvanometer (80 Ω) from the total resistance gives the series resistance required: $100,000 \Omega - 80 \Omega = 99,920 \Omega$, which is approximately $100 k\Omega$.

A Galvanometer with a Coil Resistance Of 80Ω80 \Omega80Ω Deflects Full-Scale for a Current Of

A Galvanometer with a Coil Resistance Of $80 \Omega$ Deflects Full-Scale for a Current Of