JEE Mains · Physics · STD 12 - 3. current electricity
Two resistance of \(100\ \Omega\) and \(200\ \Omega\) are connected in series with a battery of \(4 \mathrm{~V}\) and negligible internal resistance. \(A\) voltmeter is used to measure voltage across \(100 \Omega\) resistance, which gives reading as \(1 \mathrm{~V}\). The resistance of voltmeter must be _______ \(\Omega\).
- A \(100\)
- B \(200\)
- C \(300\)
- D \(400\)
Answer & Solution
Correct Answer
(B) \(200\)
Step-by-step Solution
Detailed explanation
\(\frac{R_V \cdot 100}{R_\gamma+100}=\frac{200}{3}\) \(3 R_v=2 R_v+200\) \(R_v=200\)
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