Demonstrative Video
Problem-1
Let \(L_1 = 0.4~\mathrm{H}\), \(L_2 = 2.5~\mathrm{H}\), \(k=0.6\), and \(i_1= 4i_2 = 20\cos(500t-20^{\circ})~\mathrm{mA}\) . Determine \(v_1(0)\) and the total energy stored in the system at \(t=0\).
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
\[\begin{aligned}
M & = k\sqrt{L_1L_2} = 0.6 \sqrt{40\times 5} =
8.4853~\mathrm{mH} \\
40~\mathrm{mH} & \Rightarrow j2000 \times 40 \times 10^{-3}
= j80 \\
5~\mathrm{mH} & \Rightarrow j2000 \times 5 \times 10^{-3} =
j10 \\
8.4853~\mathrm{mH} & \Rightarrow j2000 \times 8.4853 \times
10^{-3} = j16.97
\end{aligned}\]
\[\begin{aligned}
V_1 & = j80I_1-j16.97I_2\\
V_2 & = -16.97I_1+j10I_2
\end{aligned}\]
\[\begin{aligned}
\text{Given} &
\begin{cases}
V_1 & = 10 \angle 0^{\circ} \\
I_2 & = 2\angle -90^{\circ}
\end{cases}
\end{aligned}\]
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:
Problem-5
Determine \(L_{eq}\)
Solution-5
\[\begin{aligned}
\text{Let}~\omega & = 1~\mathrm{rad/sec} \\
\text{KVL in Loops} & \\
1 & = j8I_1 +j4I_2 \\
0 & = j4I_1 +j18I_2 \\
\text{Solving} & \\
I_1 & = -j0.1406 \\
Z & = \dfrac{1}{I_1} = jL_{eq} \\
\Rightarrow~L_{eq} & = \dfrac{1}{jI_1} = 7.111~\mathrm{H}
\end{aligned}\]
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
\[\begin{aligned}
\text{Given}~~j\omega L & = j40 \qquad
\Rightarrow~\omega = \dfrac{40}{L} = \dfrac{40}{15\times10^{-3}} =
2667~\mathrm{rad/sec} \\
M & = k \sqrt{L_1L_2} = 0.6 \sqrt{12 \times 10^{-3}\times
30\times 10^{-3}} = 11.384~\mathrm{mH}\\
\end{aligned}\]
\[\begin{aligned}
15~\mathrm{mH} & \Rightarrow 40~\Omega \\
12~\mathrm{mH} & \Rightarrow 32~\Omega \\
30~\mathrm{mH} & \Rightarrow 80~\Omega \\
11.384~\mathrm{mH} & \Rightarrow 30.36~\Omega \\
\end{aligned}\]
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
\[\begin{aligned}
M & = k\sqrt{L_1L_2} \\
\Rightarrow~\omega M & = k\sqrt{\omega L_1 \omega L_2} =
0.5\times 10 = 5\\
\text{Mesh-1} & \quad j12 = (4+j5)I_1 +j10I_2 \\
\text{Mesh-2} & \quad -20 = j10I_1 + (8+j5)I_2 \\
\text{Solving} & \\
I_1 & = 2.462\angle 72.18^{\circ} \\
I_2 & = 0.878\angle -97.48^{\circ} \\
I_3 & = I_1-I_2 = 3.329\angle 74.89^{\circ}
\end{aligned}\]
\[\begin{aligned}
i_1(t) & = 2.462\cos(1000t+72.18^{\circ})~\mathrm{A} \\
i_2(t) & = 0.878\cos(1000t-97.48^{\circ})~\mathrm{A} \\
t &= 2~\mathrm{ms},~1000t = 2~\mathrm{rad} = 114.6^{\circ}\\
i_1(t=2~\mathrm{ms}) & = 2.462\cos(114.6+72.18) = -2.445 \\
i_2(t=2~\mathrm{ms}) & = 0.878\cos(114.6-97.48) = -0.8391\\
\omega \mathrm{L}_{1}&=10 \text { and } \omega=1000\\
\Rightarrow & \mathrm{~L}_{1}=\mathrm{L}_{2}=10 ~\mathrm{mH}\\
\Rightarrow & \mathrm{M}=0.5 \mathrm{L}_{1}=5 \mathrm{mH}\\
W & = 0.5L_1i_1^2+0.5L_2i_2^2-Mi_1i_2\\
& = 43.67~\mathrm{mJ}
\end{aligned}\]
Problem-10
Calculate the total inductance of the there-coupled coils.
Solution-10
\[\begin{aligned}
\text{Coil-1} & \quad L_{1}-M_{12}+M_{13}=6-4+2=4 \\
\text{Coil-2} & \quad L_{2}-M_{21}-M_{23}=8-4-5=-1 \\
\text{Coil-3} & \quad L_{3}+M_{31}-M_{32}=10+2-5=7 \\
L_T & = 4-1+7 = 10~\mathrm{H}
\end{aligned}\]
\[\begin{aligned}
L_T & = L_1+L_2+L_3-2M_{12}-2M_{23}+2M_{12}\\
& = 10~\mathrm{H}
\end{aligned}\]
OR
Problem-11
Calculate \(i_2(t)\) if \(v_1(t)=8\sin720t\)
Solution-11
\(\omega = 720~\mathrm{rad/sec}\)
Mesh equations:
\[\left[\begin{array}{ccc}
1.8+j 0.72 & -j 0.72036 & j 3.6 \times 10^{-4} \\
-j 0.72036 & j 0.77183 & -j 1.44036 \\
j 3.6 \times 10^{-4} & -j 1.44036 & 2+j 1.44
\end{array}\right]\left[\begin{array}{l}
\mathbf{I}_{1} \\
\mathbf{I}_{2} \\
\mathbf{I}_{3}
\end{array}\right]=\left[\begin{array}{l}
8 \\
0 \\
0
\end{array}\right]\]