AP EAMCET · PHYSICS · Current Electricity
A maximum current of 0.5 mA can pass through a galvanometer of resistance \(15 \Omega\). The resistance to be connected in series to the galvanometer to convert it into a voltmeter of range \(0-10 \mathrm{~V}\) is
- A \(9985 \Omega\)
- B \(20015 \Omega\)
- C \(20000 \Omega\)
- D \(19985 \Omega\)
Answer & Solution
Correct Answer
(D) \(19985 \Omega\)
Step-by-step Solution
Detailed explanation
\(V = I_g (R_g + R_s)\) \(10 = (0.5 \times 10^{-3}) (15 + R_s)\) \(R_s = \frac{V}{I_g} - R_g = \frac{10}{0.5 \times 10^{-3}} - 15\) \(R_s = 20000 - 15 = 19985 \, \Omega\)
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