The switch in the circuit of Fig. 30-15 has been closed on a for a very long time when it is then thrown to b. The resulting current through the inductor is indicated in Fig. 30-28 for four sets of values for the resistance R and inductance L: (1) $\mathit{R}\mathbf{=}{\mathit{R}}_{\mathbf{0}}$, (2) $\mathbf{2}{\mathit{R}}_{\mathbf{0}}\mathbf{=}\mathit{R}$, (3) $\mathit{R}\mathbf{0}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{2}\mathit{L}\mathbf{0}$ , (4) $\mathbf{2}\mathit{R}\mathbf{0}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{2}\mathit{L}\mathbf{0}$. Which set goes with which curve?

1. Fig. 30-28.
2. Fig.30-15.
3. The switch has been closed on ‘a’ for a very long time when it is then through to ‘b’.
4. In set (1)$R=R_0$and$L=L_0$.
5. In set (2)$R=2R_0$and$L=L_0$.
6. In set (3)$R=R_0$androle="math" localid="1661834742211" $L=2L_0$.
7. In set (4) $R=2R_0$ and $L=2L_0$.

Applying Ohm’s law and using Eq.30-42, it can find the induced current i and the inductive time constant ${\mathit{\tau }}_{\mathbf{L}}$for corresponding given sets. Comparing these values with the given curves in Fig.30-28, find which curve indicates which set.

Formulae are as follow:

$i=\frac{{\epsilon }_{ind}}{R}$.

$i=\frac{\epsilon }{R}\left(1-e^{-t/T_L}\right)$

${\tau }_L=\frac{L}{R}$

From Fig.30-28, the current decreases exponentially with respect to time .

According to Ohm’s law, the current is,

$i=\frac{{\epsilon }_{ind}}{R}...............................................................................\left(1\right)$

Where, ${\epsilon }_{ind}$is induced emf.

From Eq.30-41, the current is,

$i=\frac{\epsilon }{R}\left(1-e^{-t/T_L}\right)...............................................................................(30-41)$

Where, ${\tau }_L$the inductive time constant and is given by,

${\tau }_L=\frac{L}{R}...............................................................................(30-42)$

In set (1) $R=R_0$and $L=L_0$, from Eq.1 and Eq.30-42, it gives,

$i_1=\frac{{\epsilon }_{ind}}{R_0}$

and

${\tau }_1=\frac{L_0}{R_0}$

In set (2) $R=2R_0$and $L=L_0$, from Eq.1 and Eq.30-42, it gives

$i_2=\frac{{\epsilon }_{ind}}{2R_0}$

and

${\tau }_2=\frac{L_0}{2R_0}$

In set (3) $R=R_0$and $L=2L_0$, from Eq.1 and Eq.30-42, it gives

role="math" localid="1661836172518" $i_3=\frac{{\epsilon }_{ind}}{R_0}$

and

${\tau }_3=\frac{2L_0}{R_0}$

In set (4) $R=2R_0$ and $L=2L_0$, from Eq.1 and Eq.30-42, it gives

$i_4=\frac{{\epsilon }_{ind}}{2R_0}$

and

role="math" localid="1661836360441" ${\tau }_4=\frac{2L_0}{2R_0}$

Therefore,

${\tau }_4=\frac{L_0}{R_0}$

Therefore, comparing the above solutions it gives,

$i_1=i_3>i_2=i_4$

And the inductive time constant,

${\tau }_3={\tau }_1>{\tau }_4={\tau }_2$

If the time constant has a larger value, it means decay is slower in RL circuits, and there is fast decay in RL circuit when the time constant has smaller value. Considering this in the above solutions and from Fig.30-28, the curves it gives

$c>a\approx d>b$.

Hence, from the above results and from Fig.30-28,

(1) curve a indicates set 2, (2)curve b indicates set 4, (3) curve c indicates set 1, and (4) curve d indicates set 3.

Using Ohm’s law and Eq.30-42, find the answer to this question.