Question

It has been proposed to use large inductors as energy storage devices. (a) How much electrical energy is converted to light and thermal energy by a$\mathbf{150}\mathit{W}$light bulb in one day? (b) If the amount of energy calculated in part (a) is stored in an inductor in which the current is$\mathbf{80}\mathit{A}$, what is the inductance?

Short answer

a) $U=1.296\times {10}^{7}J$

b)$L=4.05\times {10}^{3}H$

Step-by-step solution

Formulas used in the given problem

The energy stored in the inductor is given by $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}{\mathit{I}}^{\mathbf{2}}$.

The power and time period are also given whose product will give us the energy. Use the formula of energy to calculate the self-inductance.

Calculate the self-inductance

Given that $P=150W,I=80A$and $t=24h$. The energy is given by

$$\begin{gathered} U=Pt \\ U=150\times 24\times 3600 \\ U=1.296\times {10}^{7}J \end{gathered}$$

Use the value of energy to find the self-inductance

localid="1664179296238" $$\begin{gathered} L=\frac{2U}{I^2} \\ L=\frac{2\times 1.296\times {10}^7}{{80}^2} \\ L=4.05\times {10}^{3}H \end{gathered}$$