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Two identical resistors each of resistance 10 ohm are connected in(i) series, (ii) parallel to a battery of 6V. Calculate the ratio of power consumed by the combination of resistor in the two cases. |
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Answer» Solution :`R_(1) = R_(2)=10 Omega` (i) When connected in series `R_(s)=R_(1)+R_(2)= 20 Omega` (ii) When connected in PARALLEL `R_(p)= (R_(1)R_(2))/(R_(1)+R_(2))= (10xx10)/(10+10)=5 Omega` `P_(p)= (V^(2))/(R_(p))=(6xx6)/(5) = 7.2 ` W `(P_(s))/(P_(p))= (1.8W)/(7.2W)=(1)/(4)= 1:4` |
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