Question

An LR circuit with emf $\epsilon$ is connected at t = 0.
(a) Find the charge Q which flows through the battery during 0 to t.
(b) Calculate the work done by the battery during this period.
(c) Find the heat developed during this period
(d) Find the magnetic field energy stored in the circuit at time t.
(e) Verify that results in the three parts above are consistent with energy conservation.

Answer 1

Emf $=E$ LR circuit
(a) $dq=idt=i_0(1-e^{-t/\tau })=i_0(1-e^{-IR/L})[\because \tau =L/R]$
$Q={\int }_0^{t}dq=i_0[{\int }_0^{t}dt-{\int }_0^t{e}^{-tR/L}dt]=i_0[t-(-L/R\left)\left(e^{-tR/L}\right)t_0\right]$
$Q=E/R[t-L/R\left({1-e}^{-tR/L}\right)]$
(b) Similarly as we know work done $=VI=EI$
$=E{i}_0[t-L/R\left({1-e}^{-tR/L}\right)]=\frac{E^2}{R}[t-L/R\left({1-e}^{-tR/L}\right)]$
(c) $H={\int }_0^t{i}^{2}Rdt=\frac{E^2}{R^2}.R.{\int }_0^t{\left({1-e}^{-tR/L}\right)}^{2}dt=\frac{E^2}{R^2}{\int }_0^t(1+e^{(-2+B)/L}-2e^{-tR/L})dt$
$=\frac{E^2}{R^2}{(t-\frac{L}{2R}{e}^{-2tR/L}+\frac{L}{R}2.e^{-2tR/L})}_0^t=\frac{E^2}{R^2}(t-\frac{L}{2R}{e}^{-2tR/L}+\frac{2L}{R}{e}^{-2tR/L})-(-\frac{L}{2R}+\frac{2L}{R})$
$=\frac{E^2}{R^2}[(t-\frac{L}{2R}{x}^2+\frac{L}{2R}x)-\frac{3L}{2R}]=\frac{E^2}{2}[t-\frac{L}{2R}(x^2-4x+3)]$
(d) $E=\frac{1}{2}L{i}^2=\frac{1}{2}L.\frac{E^2}{R^2}{\left({1-e}^{-tR/L}\right)}^2[x=e^{-tR/L}]=\frac{{LE}^2}{{2R}^2}{(1-x)}^2$
(e) Total energy used as heat as stored in magnetic field
$=\frac{E^2}{R}T-\frac{E^2}{R}.\frac{L}{2R}{x}^2+\frac{E^2}{R}\frac{L}{r}.4x^2-\frac{3L}{2R}.\frac{E^2}{R}+\frac{{LE}^2}{R^2}+\frac{{LE}^2}{{2R}^2}{x}^2-\frac{{LE}^2}{R^2}x=\frac{E^2}{R}(t-\frac{L}{R}(1-x\left)\right)$
$=$ Energy drawn from battery