InterviewSolution
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InterviewSolution
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(a) The potential difference applied across a given resistor is altered so that the heat produced per secound increases by a factor of 9 . By what factor does the appliced potential difference change ? (b) In the figure shown, an ammeter A and a resistor of 4 `Omega` are connected to the terminals of the socrce . The emf of the source is 12 V having an internal resisitance of 2 `Omega` Calculate the voltmeter and ammeter readings. |
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Answer» (a) H = Vlt ` H_(1) = (V_(1)^(2) t)/(R ) and H_(2) = (V_(2)^(2) )/(R )t` Taking `H_(2) =9H_(1),(H_(1))/(H_(2))= (H_(1))/(9H_(1) )= ((V_1^2)/cancel Rcancelt)/((V_2^2)/cancelRcancel t)` `rArr (1)/(9) = (V_(1)^(2))/(V_(1)^(2) )=((V_(1))/(V_(2) ))^(2) rArr (V_(1))/(V_(2) )= (1)/(3)` `rArr V_(2) 3V_(1)` Hence, Increase by a factor of 3. (b) Ammeter reading `I = (E)/((r + R)) = (12) /(2+4) = (12)/(6) = 2 Amp` Volmeters reading = V = E - Ir = 12 - (2.2)V = 8V. |
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