Chapter 32, Problem 30P

(a)

To determine

To show: The equivalent inductance is $L_{\text{eq}}=L_1+L_2$.

Answer to Problem 30P

The equivalent inductance in series combination is $L_{\text{eq}}=L_1+L_2$.

Explanation of Solution

Given info: The inductance of the inductor are $L_1$ and $L_2$. The internal resistances of them are zero.

For a series connection, both inductor carry equal currents at every instant. So the change in current $\frac{di}{dt}$ is same for both inductor.

Write the expression to calculate the voltage across the pair.

$\begin{array}{c}{L}_{\text{eq}}\frac{di}{dt}=L_1\frac{di}{dt}+L_2\frac{di}{dt}\\ L_{\text{eq}}=L_1{+}_{L2}\end{array}$

Conclusion:

Therefore, the equivalent inductance is $L_{\text{eq}}=L_1+L_2$.

(b)

To determine

To show: The equivalent inductance in parallel combination is $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$.

Answer to Problem 30P

The equivalent inductance in parallel combination is $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$.

Explanation of Solution

Given info: The inductance of the inductor are $L_1$ and $L_2$. The internal resistances of them are zero.

For a parallel connection, the voltage across each inductor is same for both.

Write the expression to calculate the voltage across each inductor.

$L_{\text{eq}}\frac{di}{dt}=L_2\frac{di}{dt}=L_2\frac{di}{dt}=\Delta V_{\text{L}}$

The current in the connection is,

$\frac{di}{dt}=\frac{d{i}_1}{dt}+\frac{d{i}_2}{dt}$ (1)

Here,

$i_1$ is the current in first inductor.

$i_2$ is the current in second inductor.

The change in current in equivalent inductor is,

$\frac{di}{dt}=\frac{\Delta V_{\text{L}}}{L_{\text{eq}}}$

The change in current in first inductor is,

$\frac{d{i}_1}{dt}=\frac{\Delta V_{\text{L}}}{L_1}$

The change in current in second inductor is,

$\frac{d{i}_2}{dt}=\frac{\Delta V_{\text{L}}}{L_2}$

Substitute $\frac{\Delta V_{\text{L}}}{L_{\text{eq}}}$ for $\frac{di}{dt}$, $\frac{\Delta V_{\text{L}}}{L_1}$ for $\frac{d{i}_1}{dt}$ and $\frac{\Delta V_{\text{L}}}{L_2}$ for $\frac{d{i}_2}{dt}$ in equation (1).

$\begin{array}{c}\frac{\Delta V_{\text{L}}}{L_{\text{eq}}}=\frac{\Delta V_{\text{L}}}{L_1}+\frac{\Delta V_{\text{L}}}{L_2}\\ \frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}\end{array}$

Thus, the equivalent inductance in parallel combination is $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$.

Conclusion:

Therefore, the equivalent inductance in parallel combination is $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$.

(c)

To determine

To show: The equivalent inductance and resistance when their internal resistance is non zero in series combination is $L_{\text{eq}}=L_1+L_2$ and $R_{\text{eq}}=R_1+R_2$ respectively.

Answer to Problem 30P

The equivalent inductance in parallel combination is $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$.

Explanation of Solution

Given info: The inductance of the inductor are $L_1$ and $L_2$. The internal resistances of them are $R_1$ and $R_2$.

Write the expression to calculate the voltage across the connection.

$L_{\text{eq}}\frac{di}{dt}+R_{\text{eq}}i=L_1\frac{di}{dt}+i{R}_1+L_2\frac{di}{dt}+i{R}_2$

Here, $i$ and $\frac{di}{dt}$ are independent quantities under our control so functional equality requires both $L_{\text{eq}}=L_1+L_2$ and $R_{\text{eq}}=R_1+R_2$ conditions satisfied.

Thus, the equivalent inductance and resistance when their internal resistance is non zero in series combination is $L_{\text{eq}}=L_1+L_2$ and $R_{\text{eq}}=R_1+R_2$ respectively.

Conclusion:

Therefore, the equivalent inductance and resistance when their internal resistance is non zero in series combination is $L_{\text{eq}}=L_1+L_2$ and $R_{\text{eq}}=R_1+R_2$ respectively.

(d)

To determine

Whether it is necessarily true that they are equivalent to a single ideal inductor having $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$ and $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$.

Answer to Problem 30P

It is necessarily true that they are equivalent to a single ideal inductor having $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$ and $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$.

Explanation of Solution

Given info: The inductance of the inductor are $L_1$ and $L_2$. The internal resistances of them are $R_1$ and $R_2$.

If the circuit elements are connected in parallel then two conditions always true.

The first condition is,

$\begin{array}{c}i=i_1+i_2\\ \frac{\Delta V_{\text{L}}}{R_{\text{eq}}}=\frac{\Delta V_{\text{L}}}{R_1}+\frac{\Delta V_{\text{L}}}{R_2}\\ \frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}\end{array}$

The second condition is,

$\begin{array}{c}\frac{di}{dt}=\frac{d{i}_1}{dt}+\frac{d{i}_2}{dt}\\ \frac{\Delta V_{\text{L}}}{L_{\text{eq}}}=\frac{\Delta V_{\text{L}}}{L_1}+\frac{\Delta V_{\text{L}}}{L_2}\\ \frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}\end{array}$

Thus, it is necessarily true that they are equivalent to a single ideal inductor having $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$ and $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$.

Conclusion:

Therefore, it is necessarily true that they are equivalent to a single ideal inductor having $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$ and $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$.