The load impedance connected to the secondary winding of the ideal transformer in Fig. 9.45 consists of a 237.5 m Ω resistor in series with a 125 μH inductor. If the sinusoidal voltage source ( vg)is generating the voltage 2500 cos 400t V, find the steady-state expressions for:(a) i1 ; (b) v1; (c)

Question

The load impedance connected to the secondary winding of the ideal transformer in Fig. 9.45 consists of a 237.5 m Ω resistor in series with a 125 μH inductor. If the sinusoidal voltage source [latex]\left( v_{g}\right)[/latex] is generating the voltage [latex]2500\cos 400t V[/latex], find the steady-state expressions for:(a)[latex] i_{1}[/latex] ; (b)[latex] v_{1}[/latex] ; (c)[latex] i_{2}[/latex] ; and (d)[latex] v_{2}[\latex];

Accepted Answer

a) We begin by constructing the phasor domain equivalent circuit. The voltage source becomes [latex]2500\angle 0^{\circ}[/latex] V; the 5 mH inductor converts to an impedance of j2 Ω and the 125 μH inductor converts to an impedance of j0.05 Ω . The phasor domain equivalent circuit is shown in Fig. 9.46.
It follows directly from Fig. 9.46 that

[latex]2500\angle 0^{\circ}= (0.25 + j 2)\pmb{I}_{1} +\pmb{V}_{1}[/latex],

and

[latex]\pmb {V}_{1}= 10\pmb {V}_{2} = 10\left[\left(0.2375 + j0.05\right)\pmb{I}_{2}\right][/latex].

Because

[latex]\pmb{I}_{2}=10\pmb{I}_{1}[/latex]

we have

[latex]\pmb {V}_{1}= 10(0.2375 + j0.05)10\pmb{I}_{1}[/latex]

[latex]= \left(23.75 + j5\right)\pmb{I}_{1}[/latex].

Therefore

[latex]2500\angle 0^{\circ}= (24 + j 7)\pmb{I}_{1}[/latex],

or

[latex]\pmb {I}_{1}= 100\angle-16.26^{\circ} A[/latex].

Thus the steady-state expression for [latex] i_{1}[/latex] is

[latex] i_{1}=100 \cos \left(400t - 16.26^{\circ}\right) A.[/latex]

b)[latex]\pmb {V}_{1}= 2500\angle 0° - \left(100 \angle -16.26°\right )\left(0.25 + j 2\right)[/latex]

= 2500 - 80 - j185

[latex]= 2420 - j185 = 2427.06 \angle -4.37° V[/latex].

Hence

[latex]v_{1}= 2427.06 \cos \left(400t - 4.37^{\circ}\right) V[/latex].

c) [latex]\pmb{I}_{2}= 10\pmb {I}_{1} =1000\angle-16.26^{\circ} A [/latex].

Therefore

[latex]i_{2} = 1000 \cos \left(400t - 16.26^{\circ}\right) A[/latex].

d)[latex]\pmb {V}_{2}= 0.1\pmb {V}_{1} = 242.71 \angle -4.37^{\circ}[/latex] V,

giving

[latex]v_{2} = 242.71 \cos \left(400t - 4.37^{\circ}\right) V[/latex].

Question 9.14: The load impedance connected to the secondary winding of the...

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