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A coil, capacitor and an AC source of r.m.s voltage 24 V are connected in series. By varying the frequency of the source, a maximum r.m.s current of 6A is observed. If this coil is connected to a battery of e.m.f 12 V and internal resistance $4 Ω$ , the current through it will be $15 \times 10^{- P} A$ . what is the value of ‘P’?

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Detailed Solution

${(I_{rms})}_{\max} = \frac{ε_{rms}}{R} \Rightarrow R = \frac{ε_{rms}}{{(I_{rms})}_{\max}} = \frac{24}{6} = 4 Ω$ .

When coil is connected across DC source,

$I = \frac{E}{R + r} = \frac{12}{4 + 4} = 1 .5 = 15 \times 10^{- 1} A$ .

$∴ P = 1$ .

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