A coil of inductance 2H and resistance 10Ω are in a series circuit with an open key and a cell of constant 100V with negligible resistance. At time t=0, the key is closed. Find the current in the circuit at t=1s
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Solution
Given : R=10ΩL=2HV=100 volts Current flowing as a function of time is given by I=VR(1−e−RtL) ∴I=10010(1−e−10t2) Or I=10(1−e−5t)A Current flowing at t=1s, ∴I=10(1−e−5×1) Or I=10(1−e−5)=10(1−6.73×10−3)=9.93A