Sol. Let the currents through inductive branch and capacitive branch be I L and I C respectively. Then the current through he battery, from KCL, is

I = I L + I C

Since the battery is connected in parallel to the RL and RC branches of the circuit, the current in the RL branch is unaffected by the presence of the branch; so

I L = [1 – e –(R/L)t ]

and I C = e –(t/Rc) Hence the current through battery is

I = [1 – e

–(R/L)t + e –t/R/C ]

If the current has to reach its final value E/R instantaneously, the exponential terms must cancel out, i.e.,

e –t/ τ L = e –t/ τ C which is possible if τ L = τ C

L/R = RC

R =

For all t > 0, this is the desired result.