An inductor of inductance L and resistance R is connected to a battery of EMF E . After a long time, the current in the circuit reaches a steady state. What is the total energy supplied by the battery

Question

An inductor of inductance L and resistance R is connected to a battery of EMF E . After a long time, the current in the circuit reaches a steady state. What is the total energy supplied by the battery during the time the current grows from zero to its steady-state value?

TestHub · Accepted Answer

The steady-state current in the L-R circuit is $I_{0} = E / R$ .

The energy balance equation for an L-R circuit is $E I = I^{2} R + L I \frac{d I}{d t}$ . Integrating over time from $0$ to $\infty$ (steady state): $\int_{0}^{\infty} E I d t = \int_{0}^{\infty} I^{2} R d t + \int_{0}^{\infty} L I \frac{d I}{d t} d t$ .

The first term is the total energy supplied by the battery ( $U_{ba tt}$ ). The second term is the total energy dissipated in the resistor ( $U_{R}$ ). The third term is the total energy stored in the inductor ( $U_{L} = \frac{1}{2} L I_{0}^{2}$ ).

For an L-R circuit, it is a known result that the energy dissipated in the resistor during charging is equal to the energy stored in the inductor, i.e., $U_{R} = \frac{1}{2} L I_{0}^{2}$ . Thus, $U_{ba tt} = U_{R} + U_{L} = \frac{1}{2} L I_{0}^{2} + \frac{1}{2} L I_{0}^{2} = L I_{0}^{2} = L (E / R)^{2} = \frac{L E ^{2}}{R ^{2}}$ .

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