JEE Mains · Physics · STD 12 - 3. current electricity
Two resistors \(400\, \Omega\) and \(800\, \Omega\) are connected in series across a \(6 V\) battery. The potential difference measured by a voltmeter of \(10\, k \Omega\) across \(400\, \Omega\) resistor is close to\(....V\)
- A \(2\)
- B \(1.95\)
- C \(2.05\)
- D \(1.8\)
Answer & Solution
Correct Answer
(B) \(1.95\)
Step-by-step Solution
Detailed explanation
So the potential difference in voltmeter across the points \(A\) and \(B\) is \(\frac{6}{1185} \times 385=1.949 V\)
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