Question

In Figure,$\mathit{R}\mathbf{=}\mathbf{15}\mathit{\Omega }\mathbf{,}\mathit{L}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{H}$the ideal battery has $\mathit{\epsilon }\mathbf{=}\mathbf{10}\mathit{V}$, and the fuse in the upper branch is an ideal3.0 A fuse. It has zero resistance as long as the current through it remains less than3.0 A . If the current reaches3.0 A , the fuse “blows” and thereafter has infinite resistance. Switch S is closed at timet = 0 . (a) When does the fuse blow? (Hint: Equation 30-41 does not apply. Rethink Eq. 30-39.) (b) Sketch a graph of the current i through the inductor as a function of time. Mark the time at which the fuse blows.

Short answer

a) Time at which the fuse blows is t = 1.5 s

b) The sketch of graph of the current through the inductor as a function of time and the time at which fuse blows is given below.

Step-by-step solution

Given

i) Resistance is $R=15\Omega$

ii) Inductance is $L=5.0H$

iii) Battery emf is $\epsilon =10V$

iv) Current through the fuse is $i=3.0A$

Understanding the concept

We use the concept of loop rule for the give circuit. Applying the loop rule, we can find the time at which the fuse blows. We can sketch the graph of the current through the inductor as a function of time.

Formulae:

For the given loop,

$\underset{}{\overset{}{\sum V=0}}$

(a) Calculate the time at which the fuse blows.

Time at which the fuse blows:

We can apply the loop rule.

$$\begin{gathered} \epsilon -L\frac{di}{dt}-iR=0 \\ \epsilon =L\frac{di}{dt}+iR \end{gathered}$$

Before the fuse blows, the current through the resistor is zero, so we can write

role="math" localid="1661427781225" $$\begin{gathered} \epsilon =L\frac{di}{dt} \\ dt=\frac{L}{\epsilon }di \end{gathered}$$

Integrating on both sides, we get

$$\begin{gathered} \int dt=\frac{L}{\epsilon }\int di \\ t=\frac{L}{\epsilon }i \end{gathered}$$

Plugging the values, we get

$$\begin{gathered} t=\frac{5.0}{10}\left(3.0\right) \\ t=1.5s \end{gathered}$$

(b) Sketch the graph of the current through the inductor as a function of time and mark the time at which the fuse blows: