A 14.0-V battery is connected in series with a switch,5.0-Ω resistor and coil.What energy is stored in the coil when the current is 2.8 A? The time constant is 4.8 × 10−⁴ s. A)9.4E-3 J B)1.9E-2 J C)6.7E-3 J D)3.8E-3 J E)1.2E-3 J

Question

Quizplus · Accepted Answer

The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2}LI^2$, where $E$ is the energy in joules, $L$ is the inductance in henries, and $I$ is the current in amperes. The time constant ($\tau$) of an RL circuit is given by $\tau = \frac{L}{R}$, where $R$ is the resistance in ohms. Rearranging this formula to solve for $L$ gives $L = \tau R$. Substituting the given values, $L = 4.8 \times 10^{-1} \times 5.0 = 2.4$ H. Substituting $L = 2.4$ H and $I = 2.8$ A into the energy formula gives $E = \frac{1}{2} \times 2.4 \times (2.8)^2 = 9.4 \times 10^{-3}$ J.

A 140-V Battery Is Connected in Series with a Switch,5