As per the given question,
Number of turns in the primary coil( N 1 ) = 500 (N_1)=500 ( N 1 ) = 500
Number of turns in the secondary coil ( N 2 ) = 200 (N_2)=200 ( N 2 ) = 200 =200
Resistance in the primary coil ( R 1 ) = 0.3 Ω (R_1)=0.3\Omega ( R 1 ) = 0.3Ω
Resistance in the secondary coil( R 2 ) = 0.02 Ω (R_2)=0.02\Omega ( R 2 ) = 0.02Ω
Leakage reactance in the primary coil ( X 1 ) = 2 Ω (X_1)=2\Omega ( X 1 ) = 2Ω
Leakage reactance in the secondary coil ( X 2 ) = 0.05 Ω (X_2)=0.05\Omega ( X 2 ) = 0.05Ω
Let R 1 ′ R_1' R 1 ′ be the resistance of the resistance of primary referred to as secondary ,
⇒ R 1 ′ = R 1 ( N 2 N 1 ) 2 \Rightarrow R_1' =R_1(\frac{N_2}{N_1})^2 ⇒ R 1 ′ = R 1 ( N 1 N 2 ) 2
⇒ R 1 ′ = 0.3 × ( 200 500 ) 2 = 0.3 × 4 25 Ω = 0.048 Ω \Rightarrow R_1'=0.3\times (\frac{200}{500})^2=\frac{0.3\times 4}{25}\Omega=0.048\Omega ⇒ R 1 ′ = 0.3 × ( 500 200 ) 2 = 25 0.3 × 4 Ω = 0.048Ω
be the resistance of the secondary referred to as primary,
⇒ R 2 ′ = R 2 ( N 1 N 2 ) 2 = 0.02 × ( 500 200 ) 2 Ω \Rightarrow R_2'=R_2(\frac{N_1}{N_2})^2=0.02\times (\frac{500}{200})^2\Omega ⇒ R 2 ′ = R 2 ( N 2 N 1 ) 2 = 0.02 × ( 200 500 ) 2 Ω
= 0.02 × 6.28 Ω = 0.1248 Ω =0.02\times 6.28 \Omega =0.1248\Omega = 0.02 × 6.28Ω = 0.1248Ω
Let X 1 ′ X_1' X 1 ′ be the leakage reactance of the primary referred to as secondary,
⇒ X 1 ′ = X 1 ( N 2 N 1 ) 2 = 2 × ( 200 500 ) 2 = 8 25 Ω = 0.32 Ω \Rightarrow X_1'=X_1(\frac{N_2}{N_1})^2=2\times (\frac{200}{500})^2=\frac{8}{25}\Omega=0.32\Omega ⇒ X 1 ′ = X 1 ( N 1 N 2 ) 2 = 2 × ( 500 200 ) 2 = 25 8 Ω = 0.32Ω
Let X 2 ′ X_2' X 2 ′ be the leakage reactance of the secondary referred to as primary,
⇒ X 2 ′ = X 2 ( N 1 N 2 ) 2 = 0.05 × ( 500 200 ) 2 = 0.05 × 25 4 = 0.3125 Ω \Rightarrow X_2' =X_2(\frac{N_1}{N_2})^2=0.05\times (\frac{500}{200})^2=0.05\times\frac{25}{4}=0.3125\Omega ⇒ X 2 ′ = X 2 ( N 2 N 1 ) 2 = 0.05 × ( 200 500 ) 2 = 0.05 × 4 25 = 0.3125Ω
a) Equivalent resistance and reactance referred to as primary,
R e q 1 = R 1 + R 2 ′ = ( 0.3 + 0.1248 ) Ω = 0.4248 Ω R_{eq1}=R_1+R_2'=(0.3+0.1248)\Omega =0.4248\Omega R e q 1 = R 1 + R 2 ′ = ( 0.3 + 0.1248 ) Ω = 0.4248Ω
X e q 1 = X 1 + X 2 ′ = 2 Ω + 0.3125 Ω X_{eq1}=X_1+X_2'=2\Omega+0.3125\Omega X e q 1 = X 1 + X 2 ′ = 2Ω + 0.3125Ω =2.3125\Omega
b) Equivalent resistance and reactance referred to as secondary coil,
R e q 2 = R 2 + R 1 ′ = ( 0.02 + 0.048 ) Ω = 0.068 Ω R_{eq2}=R_2+R_1'=(0.02+0.048)\Omega =0.068\Omega R e q 2 = R 2 + R 1 ′ = ( 0.02 + 0.048 ) Ω = 0.068Ω
X e q 2 = X 2 + X 1 ′ = 0.05 Ω + 0.32 Ω = 0.37 Ω X_{eq2}=X_2+X_1'=0.05\Omega+0.32\Omega =0.37\Omega X e q 2 = X 2 + X 1 ′ = 0.05Ω + 0.32Ω = 0.37Ω
c) Equivalent impedance referred to as primary side,
z = R e q 1 2 + X e q 1 2 = 0.424 8 2 + 2.312 5 2 = 2.35 Ω z=\sqrt{R_{eq1}^2+X_{eq1}^2}=\sqrt{0.4248^2+2.3125^2}=2.35\Omega z = R e q 1 2 + X e q 1 2 = 0.424 8 2 + 2.312 5 2 = 2.35Ω
d) We know that output power factor = true power/apparent power
true power( P ) = E 2 R e q 2 (P)=\frac{E^2}{R_{eq2}} ( P ) = R e q 2 E 2
Apparent power ( P a ) = E 2 Z (P_a)=\frac{E^2}{Z} ( P a ) = Z E 2
output power factor=R e q 2 Z = 0.068 2.35 = 0.0289 \frac{R_{eq2}}{Z}=\frac{0.068}{2.35}=0.0289 Z R e q 2 = 2.35 0.068 = 0.0289
⇒ cos ϕ = 0.0289 \Rightarrow \cos\phi=0.0289 ⇒ cos ϕ = 0.0289
⇒ ϕ = cos − 1 ( 0.0289 ) = 88.3 4 ∘ \Rightarrow \phi=\cos^{-1}(0.0289)=88.34^\circ ⇒ ϕ = cos − 1 ( 0.0289 ) = 88.3 4 ∘
e)
f) As here the value of potential or current or power is not given so data is in sufficient to answer about the input power and output power. We can calculate it with the help of the given formula.
Apparent input power, S i n = I 2 Z = 2.35 I 2 S_{in}=I^2Z=2.35I^2 S in = I 2 Z = 2.35 I 2
Apparent output power S o u t = I 2 Z o S_{out}=I^2Z_{o} S o u t = I 2 Z o