Solved Problems on Mutual Inductance

Solution-1

\[\begin{aligned} M &=k \sqrt{L_{1} L_{2}}=0.6 \sqrt{(0.4)(2.5)}=0.6 \mathrm{H} \\ v_{1}(t) &=L_{1} \frac{d i_{1}}{d t}+M \frac{d i_{2}}{d t} \\ v_{1}(0) &=0.4\left[-10 \sin \left(-20^{\circ}\right)\right]+0.6\left[-2.5 \sin \left(-20^{\circ}\right)\right]=1.881 \mathrm{~V} \\ i_{1}(0) &=20 \cos \left(-20^{\circ}\right)=18.79 \mathrm{~mA} \\ i_{2}(0) &=0.25 i_{1}(0)=4.698 \mathrm{~mA} \\ w(t) &=\frac{1}{2} L_{1}\left[i_{1}(t)\right]^{2}+\frac{1}{2} L_{2}\left[i_{2}(t)\right]^{2}+M\left[i_{1}(t)\right]\left[i_{2}(t)\right] \\ &=151.2 \mu \mathrm{J} . \end{aligned}\]

Problem-2

• The coils have \(L_1=40\) mH, \(L_2=5\) mH, and coupling coefficient \(k=0.6\). Find \(i_1(t)\) and \(v_2(t)\), given that \(v_1(t)=10\cos\omega t\) and \(i_2(t)=2\sin\omega t\), and \(\omega = 2000\) rad/sec.

Solution-2

\[\begin{aligned} M & = k\sqrt{L_1L_2} = 0.6 \sqrt{40\times 5} = 8.4853~\mathrm{mH} \\ 40~\mathrm{mH} & \Rightarrow j2000 \times 40 \times 10^{-3} = j80 \\ 5~\mathrm{mH} & \Rightarrow j2000 \times 5 \times 10^{-3} = j10 \\ 8.4853~\mathrm{mH} & \Rightarrow j2000 \times 8.4853 \times 10^{-3} = j16.97 \end{aligned}\]

\[\begin{aligned} \text{Given} & \begin{cases} V_1 & = 10 \angle 0^{\circ} \\ I_2 & = 2\angle -90^{\circ} \end{cases} \end{aligned}\] \[\begin{aligned} I_1 & = 0.5493\angle -90^{\circ} \Rightarrow i_1(t) = 0.5493\sin\omega t\\ V_2 & = 22.0656\angle 24.99^{\circ} \Rightarrow v_2(t) = 22.065\cos(\omega t +25^{\circ}) \end{aligned}\]

Problem-3

• Find \(V_0\) marked in the circuit?

Solution-3

\[\begin{aligned} \text{Mesh-1}~~12 & = I_1(2+j6)+jI_2 \\ \text{Mesh-2}~~0 & = jI_1 + (2-j1+j4)I_2 \end{aligned}\] \[\begin{aligned} &{\left[\begin{array}{c} 12 \\ 0 \end{array}\right]=\left[\begin{array}{cc} 2+j 6 & j \\ j & 2+j 3 \end{array}\right]\left[\begin{array}{l} I_{1} \\ I_{2} \end{array}\right]} \\ &I_{2}=-0.4381+j 0.3164\\ V_0 & = I_2 \times 1 = 540.5\angle 144.16^{\circ}~~\mathrm{mV} \end{aligned}\]

Problem-4

• Find \(V_x\) marked in the network ?

Solution-4

Source-Transformation

\[\begin{aligned} \text{Mesh-1}&~~8\angle 30^{\circ} = (2+j4)I_1-jI_2 \\ \text{Mesh-2}&~~(j4+2-j)I_2-jI_1j2 = 0 \end{aligned}\] On solving: \[\begin{aligned} I_2 & = 1.037\angle 21.12^{\circ} \\ V_x & = 2I_2 = 2.074\angle 21.12^{\circ} \end{aligned}\]

Problem-5

• Determine \(L_{eq}\)

Solution-5

Problem-6

• If load is an 15 mH inductor having an impedance \(Z_L = j40~\Omega\), then determine \(Z_{in}\) when \(k = 0.6\).

Solution-6

\[\begin{aligned} \text{Given}~~j\omega L & = j40 \qquad \Rightarrow~\omega = \dfrac{40}{L} = \dfrac{40}{15\times10^{-3}} = 2667~\mathrm{rad/sec} \\ M & = k \sqrt{L_1L_2} = 0.6 \sqrt{12 \times 10^{-3}\times 30\times 10^{-3}} = 11.384~\mathrm{mH}\\ \end{aligned}\]

Problem-7

• Find \(I_0\)

Solution-7

Source Transformation

\[\begin{aligned} M & = k\sqrt{L_1L_2} \\ \omega M & = k\sqrt{\omega L_1 \omega L_2} = 0.6\sqrt{20 \times 40} = 17 \\ \text{Mesh-1} & \quad 200\angle 60^{\circ} = (50-j30+j20)I_1 - j17I_2 \\ \text{Mesh-2}& \quad 0 = (10+j40)I_2 - j17I_1\\ \text{Solving} & \\ I_2 & = I_0 =1.516 \angle 92.04^{\circ} \\ \end{aligned}\]

Problem-8

• Find the input impedance ?

Solution-8

\[\begin{aligned} \text{Mesh-1} & \quad 1 = (1+j10)I_1-j4I_2 \\ \text{Mesh-2} & \quad 0 = -jI_1+(2+j3)I_2\\ \text{Solving} & I_1 = 0.019-j0.1068 \\ Z & = \dfrac{1}{I_1} = 9.219\angle 79.91^{\circ}~\Omega \end{aligned}\]

Problem-9

• Determine \(I_1,~I_2,~I_3\) and the energy stored in the coupled coils at \(t=2~\mathrm{ms}\) taking \(\omega = 1000~\mathrm{rad/sec}\)

Solution-9

Source Transformation

\[\begin{aligned} M & = k\sqrt{L_1L_2} \\ \Rightarrow~\omega M & = k\sqrt{\omega L_1 \omega L_2} = 0.5\times 10 = 5\\ \text{Mesh-1} & \quad j12 = (4+j5)I_1 +j10I_2 \\ \text{Mesh-2} & \quad -20 = j10I_1 + (8+j5)I_2 \\ \text{Solving} & \\ I_1 & = 2.462\angle 72.18^{\circ} \\ I_2 & = 0.878\angle -97.48^{\circ} \\ I_3 & = I_1-I_2 = 3.329\angle 74.89^{\circ} \end{aligned}\]

Problem-10

• Calculate the total inductance of the there-coupled coils.

Solution-10

OR \[\begin{aligned} L_T & = L_1+L_2+L_3-2M_{12}-2M_{23}+2M_{12}\\ & = 10~\mathrm{H} \end{aligned}\]

Problem-11

• Calculate \(i_2(t)\) if \(v_1(t)=8\sin720t\)

Solution-11

\(\omega = 720~\mathrm{rad/sec}\)

Mesh equations: