The time domain behavior of an RL circuit is represented by $$$\mathrm L\frac{\mathrm{di}\left(\mathrm t\right)}{\mathrm{dt}}+\mathrm{Ri}\;=\;{\mathrm V}_0\left(1\;+\;\mathrm{Be}^{-\mathrm{Rt}/\mathrm L}\;\sin\;\mathrm t\right)\mathrm u\left(\mathrm t\right)$$$ For an initial current of i(0) = $$\frac{{\mathrm V}_0}{\mathrm R}$$, the steady state value of the current is given by

A fully charged mobile phone with a 12 V battery is good for a 10 minute talk-time. Assume that, during the talk-time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?

An AC source of RMS voltage $$20$$ $$V$$ with internal impedance $${Z_s} = \left( {1 + 2j} \right)\Omega $$ feeds a load of impedance $${Z_L} = \left( {7 + 4j} \right)\Omega $$ in the figure below. The reactive power consumed by the load is

If the transfer function of the following network is $$\,{{{V_0}(s)} \over {{V_i}(s)}} = {1 \over {2\, + \,sCR}}$$

The value of the load resistance $${R_L}$$ is