Chapter 4: Q26E (page 1014)

In Fig. $3.011$ , switch $S 1$ is closed while switch $S 2$ is kept open. The inductance is $L = 0.115 H$ , and the resistance is $R = 120 Ω$ . (a) When the current has reached its final value, the energy stored in the inductor is $0.26 J$ . What is the emf $ε$ of the battery? (b) After the current has reached its final value, $S 1$ is opened androle="math" localid="1664253376215" $S 2$ is closed. How much time does it take for the energy stored in the inductor to decrease to $0.13 J$ , half of the original value?

Expert verified

a) $ε = 256 V$

b)role="math" localid="1664253462958" $t = 3.32 \times 10^{- 4} s$

Step by step solution

The current equation is given by $i = i_{m a x} (1 - e^{- \frac{t}{T}})$ and the energy equation is given by $U = \frac{1}{2} {Li}^{2}$ .

Given that $L = 0.115 H, R = 120 Ω$ and $U = 0.26 J$ .

$U = \frac{1}{2} {Li}^{2} i = \sqrt{\frac{2 U}{L}} i = \sqrt{\frac{2 \times 0.26}{0.115}} i = 2.13 A$

Use resistance and current to find EMF.

$ε = i R ε = 2.13 \times 120 ε = 256 V$

The decreasing current is given by $i = i_{\max} e^{- \frac{t}{T}}$

.The energy stored in the inductor will be given by

$U = \frac{1}{2} {Li}^{2} U = \frac{1}{2} {Li}^{2}_{m a x} e^{- \frac{2 R}{L} t} U = U_{m a x} e^{- \frac{2 R}{L} t}$

Now, the energy becomes half the maximum value.

$\frac{U}{U_{m a x}} = \frac{1}{2} = e^{- \frac{2 R}{L} t}$

Take natural logarithm on both the sides.

localid="1664254817377" $t = - \frac{L}{2 R} \ln \frac{1}{2} t = \frac{0.115 \times \ln 2}{2 \times 120} t = 3.32 \times 10^{- 4} s$

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