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The three resistances, each of value `5 Omega` are connected to the source of emf `epsilon` through ammeter A as shown in figure. If ammeter shows a reading of 2 A, calculate the power dissipated in the circuit. |
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Answer» Total resistance of the circuit is `R = ( 5 xx 5)/( 5 + 5) + 5 = 7.5 Omega` Current in the circuit, I = 2 A Power dissipated in the circuit, `P = I^(2)R = (2)^(2) xx 7.5 = 30.0W` |
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