InterviewSolution
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InterviewSolution
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Consider the circuit shown in fig 5.174 where a battery of emf 4V and a capacitor of capacitance 1 `muF` is connected to a combination of resistacnes. Find out the stedy-state current in the battery. |
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Answer» Let us consider I to be the steady-state current through the cirucuit. In the steady-state condition , current I is constant in the circuit and the capacitor offers infinite resistance. So the resistance `10Omega` becomues ineffective in the circuit. So in this case, the euqivalent resistance of resistors `2Omega and 4Omega` that are connected in parallel is `(1)/(R_P) = (1)/(2) +(1)/(4) = (3)/(4) or R_P = (4)/(3)Omega` Therefore, the total resistance of the circuit is `(4)/(3) +(2)/(3) +(2)/(3) = (6)/(3) = 2Omega` Hence, the steady state current in the circuit is `4/2 = 2A.` |
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