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(a) The potential difference applied across a given resistor is altered so that the heat producedper second increases by a factor us 9. By what factor docs the applied potential difference change?(b) In the figure shown, an ammeter A and a resistor of 4Omega are connected to the terminals of the source. The emf of the source is 12 V having an internal resistance of 2Omega . Calculate the voltmeter and ammeter readings. |
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Answer» Solution : (a) We know that heat produced per second across a given resistor R is`H = (V^2)/(R )`,where V is the potential difference applied across it. Thus `(H.)/(H)= ( (V.)/(V))^2 `or`(V.)/(V) = sqrt( (H.)/(H)) = sqrt( (9H)/(H)) = 3 rArr V. = 3V` (b) Here `R = 4 Omega , r = 2Omega` and `E = 12 V` ` therefore` Current flowing in the circuit i.e., ammeter reading`I = (epsi)/(R+ r) = (12)/(4+2) = 2A`and VOLTMETER reading `V = E -IR = 12 - 2 xx 2 = 8V` |
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