In the given circuit, let i1 be the current drawn from battery at time t=0 i.e. when switch is just closed and i2 be steady state current at t=∞, then the ratio i1i2 is
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Solution
At t=0, inductor behave as open circuit so i1=1010=1A At t=∞, inductor behaves as a short circuit, so i2=108=54A Hence, i1i2=15/4=45=0.8