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प्रश्न
An input potential Vin = 200 sin 100 πt V is provided to an ideal transformer having 1000 turns in the primary coil and 100 turns in the secondary coil as shown in the figure. The load circuit has a resistance of 4Ω, a capacitive reactance of 2Ω, and an inductive reactance of 6Ω.

Find:
- the output voltage across the load circuit.
- the current is flowing through the load circuit.
- the power supplied to the load circuit by the transformer.
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उत्तर
Given:
Input voltage, Vin = 200 sin 100 πt V
Primary turns, Np = 1000
Secondary turns, Ns = 100
Load: R = 4 Ω, Xc = 2 Ω, XL = 6 Ω
i. Output voltage across load circuit:
`V_s= N_s/N_p V_p`
= `100/1000 xx 200 sin 100 πt V`
= 0.1 × 200 sin 100 πt V
= 20 sin 100 πt V
ii. Current flowing through load circuit:
Net reactance:
X = XL − XC
= 6 − 2
= 4 Ω
Impedance:
Z = `sqrt(R^2 + X^2)`
= `sqrt(4^2 + 4^2)`
= `sqrt(16 + 16)`
= `sqrt32`
= 5.66 Ω
RMS voltage of secondary:
Secondary output, Vs = 20 sin 100 πt
Peak value, Vm = 20 V
`V_"rms" = V_m/sqrt2`
= `20/sqrt2`
= 14.14 V
Current through the load:
I = `V_"rms"/Z`
= `14.14/5.66`
= 2.5 A
iii. Power supplied to the load:
Power factor:
`cos phi = R/Z`
= `4/5.66`
= 0.707
Real power:
P = Vrms I cos ϕ
= 14.14 × 2.5 × 0.707
= 25 W
Notes
The answers for Part (ii) and Part (iii) given in the board solution are incorrect. The correct current is 2.5 A and the correct power is 25 W.
