Question 5.30: A voltage of 115 V at a frequency of 10 kHz is applied to t......
A voltage of 115 V at a frequency of 10 kHz is applied to the primary of the coupled circuit shown in Fig. 1. Determine the total impedance referred to primary and the currents in primary and secondary.
Given that, f=10 kHz , L _p=200 \mu H , \quad C _{ p }=5 \mu F
M =75 \mu H , \quad L _{ s }=100 \mu H , \quad C _{ s }=0.1 \mu FPrimary inductive reactance = j 2 \pi fL _{ p }= j 2 \pi \times 10 \times 10^3 \times 200 \times 10^{-6}
= j12.5664 Ω
Primary capacitive reactance =\frac{1}{ j 2 \pi fC _{ p }}=\frac{1}{ j 2 \pi \times 10 \times 10^3 \times 5 \times 10^{-6}}=- j 3.1831 \Omega
Mutual reactance = j 2 \pi fM = j 2 \pi \times 10 \times 10^3 \times 75 \times 10^{-6}= j 4.7124 \Omega
Secondary inductive reactance = j 2 \pi f L _{ s }= j 2 \pi \times 10 \times 10^3 \times 100 \times 10^{-6}= j 6.2832 \Omega
Secondary capacitive reactance =\frac{1}{ j 2 \pi fC }=\frac{1}{ s 2 \pi \times 10 \times 10^3 \times 0.1 \times 10^{-6}}=- j 159.1549 \Omega
The frequency domain equivalent of the coupled circuit is shown in Fig. 2. Let \bar{I}_{ p } \text { and } \bar{I}_{ s } be the primary
and secondary currents as shown in Fig. 2. The electrical equivalent of the coupled circuit is shown in Fig. 3. (Here it is group-I coupled circuit).
With reference to Fig. 3, the mesh basis matrix equation can be obtained as shown below :
\left[\begin{array}{rr} 12+ j 7.854+ j 4.7124- j 3.1831 & – j 4.7124 \\ – j 4.7124 & j 4.7124+ j 1.5708+8- j 159.1549 \end{array}\right]\left[\begin{array}{l} \overline{ I }_{ p } \\ \overline{ I }_{ s } \end{array}\right]=\left[\begin{array}{r} 115 \angle 0^{\circ} \\0 \end{array}\right]
\left[\begin{array}{rr} 12+j 9.3833 & -j 4.7124 \\ -j 4.7124 & 8-j 152.8717 \end{array}\right]\left[\begin{array}{l} \bar{I}_p \\ \bar{I}_s \end{array}\right]=\left[\begin{array}{r} 115 \\ 0 \end{array}\right]
\text { Now, } \Delta=\left|\begin{array}{rr} 12+j 9.3833 & -j 4.7124 \\ -j 4.7124 & 8-j 152.8717 \end{array}\right|=(12+j 9.3833) \times(8-j 152.8717)-(-j 4.7124)^2
= 1552.6477 – j1759.394
\begin{aligned} \Delta_p=\left|\begin{array}{rr} 115 & -j 4.7124 \\ 0 & 8- j 152.8717 \end{array}\right| & =115 \times(8- j 152.8717)-0 \\ & =920- j 17580.2455 \end{aligned}
\begin{aligned} \Delta_{ S }=\left|\begin{array}{rr} 12+ j 9.3833 & 115 \\ – j 4.7124 & 0 \end{array}\right| & =0-(- j 4.7124) \times 115 \\ & = j 541.926 \end{aligned}
\begin{aligned} & \overline{ I }_{ p }=\frac{\Delta_{ p }}{\Delta}=\frac{920- j 17580.2455}{1552.6477- j 1759.394}=5.8769- j 4.6634 A =7.5023 \angle-38.4^{\circ} A \\ & \overline{ I }_{ s }=\frac{\Delta_{ s }}{\Delta}=\frac{ j 541.926}{1552.6477- j 1759.394}=-0.1732+ j 0.1528 A =0.231 \angle 138.6^{\circ} A \end{aligned}
Total impedance referred to primary =\frac{115 \angle 0^{\circ}}{\overline{ I }_{ p }}
=\frac{115 \angle 0^{\circ}}{7.5023 \angle-38.4^{\circ}}=15.3286 \angle 38.4^{\circ} \OmegaRESULT
\text { Primary current, } \bar{I}_p=7.5023 \angle-38.4^{\circ} A\text { Secondary current, } \overline{ I}_{ s }=0.231 \angle 138.6^{\circ} A
\text { Total impedance referred to primary }=15.3286 \angle 38.4^{\circ} A
