Chapter 32, Problem 32.30P

Two ideal inductors, L₁ and L₂, have zero internal resistance and are far apart, so their magnetic fields do not influence each other. (a) Assuming these inductors are connected in series, show that they are equivalent to a single ideal inductor having L_eq = L₁ + L₂. (b) Assuming these same two inductors are connected in parallel, show that they are equivalent to a single ideal inductor having 1/L_eq = 1/L₁ + 1/L₂. (c) What If? Now consider two inductors L₁ and L₂ that have nonzero internal resistances R₁ and R₂, respectively. Assume they are still far apart, so their mutual inductance is zero, and assume they are connected in series. Show that they are equivalent to a single inductor having L_eq = L₁ + L₂ and R_eq = R₁ + R₂. (d) If these same inductors are now connected in parallel, is it necessarily true that they are equivalent to a single ideal inductor having 1/L_eq = 1/L₁ + 1/L₂ and 1/R_eq = 1/R₁ + 1/R₂? Explain your answer.

To determine

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

Answer to Problem 32.30P

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 $L_{\text{eq}}=L_1+L_2$.

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 32.30P

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 $\frac{1}{L_{\text{eq}}}=\frac{1}{L_1}+\frac{1}{L_2}$.

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 32.30P

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.

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 32.30P

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 $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$.