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Two identical resistors, each of resistance are connected in (i) series, and (ii)parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case. |
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Answer» Solution :Here `R_(1) = R_(2) = 15 Omega` and V = 6 V (i) When the two resistors are CONNECTED in series, the EQUIVALENT resistance `R_(S)` of the CIRCUIT is. `R_(S) = R_(1) + R_(2) = 15 + 15 = 30 Omega`. `:.` Power consumed in series combination of resistors, `P_(S) = (V^(2))/(R_(S)) = (6 xx 6)/(30) = 1.2 W` When the two resistors are connected in parallel, the equivalent resistance `R_(P)` of the circuit is, `(1)/(R_(P)) = (1)/(R_(1)) +(1)/(R_(2)) = (1)/(15) + (1)/(15) =(2)/(15) rArr R_(P) = (15)/(2) = 7.5 Omega` `:.` Power consumed in parallel combination of resistors, `rArr P_(P) = (V^(2))/(R_(P)) = (6 xx 6)/(7.5) = 4.8 W rArr (P_(S))/(P_(P)) = (1.2W)/(4.8W) = (1)/(4)` |
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