JEE Mains · Physics · STD 12 - 3. current electricity
A battery of \(3.0\, V\) is connected to a resistor dissipating \(0.5\, W\) of power. If the terminal voltage of the battery is \(2.5\, V\) the power dissipated within the internal resistance is\(.......W\)
- A \(0.50\)
- B \(0.125\)
- C \(0.072\)
- D \(0.10\)
Answer & Solution
Correct Answer
(D) \(0.10\)
Step-by-step Solution
Detailed explanation
\(P_{R}=0.5 W\) \(\Rightarrow i ^{2} R =0.5 W\) \(Also , V = E - ir\) \(2.5=3-\) ir \(\Rightarrow\) ir \(=0.5\) Power dissipated across 'r' : \(P_{r}=i 2 r\) \(Now\) iR \(=2.5\) ir \(=0.5\) On dividing : \(\frac{ R }{ r }=5\)…
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