Answer to Question #123922 in Electricity and Magnetism for suzana

Question

A transformer has 500 primary turns and 200 secondary turns. The primary and secondary resistances
are 0.3Ω and 0.02Ω respectively and the corresponding leakage reactance are 2Ω and 0.05Ω
respectively. Determine:
(a) the equivalent resistance and reactance referred to the primary side
(b) the equivalent resistance and reactance referred to the primary side, impedance summed
(c) the output power factor
(d) draw the equivalent circuit referred to the primary side, impedance summed
(e) input and output apparent power

Answer 1

As per the given question,

Number of turns in the primary coil $(N_1)=500$

Number of turns in the secondary coil $(N_2)=200$

Resistance in the primary coil $(R_1)=0.3\Omega$

Resistance in the secondary coil $(R_2)=0.02\Omega$

Leakage reactance in the primary coil $(X_1)=2\Omega$

Leakage reactance in the secondary coil $(X_2)=0.05\Omega$

Let $R_1'$ be the resistance of the resistance of primary referred to as secondary ,

$\Rightarrow R_1' =R_1(\frac{N_2}{N_1})^2$

$\Rightarrow R_1'=0.3\times (\frac{200}{500})^2=\frac{0.3\times 4}{25}\Omega=0.048\Omega$

Let $R_2'$ be the resistance of the secondary referred to as primary,

$\Rightarrow R_2'=R_2(\frac{N_1}{N_2})^2=0.02\times (\frac{500}{200})^2\Omega$

$=0.02\times 6.28 \Omega =0.1248\Omega$

Let $X_1'$ be the leakage reactance of the primary referred to as secondary,

$\Rightarrow X_1'=X_1(\frac{N_2}{N_1})^2=2\times (\frac{200}{500})^2=\frac{8}{25}\Omega=0.32\Omega$

Let $X_2'$ be the leakage reactance of the secondary referred to as primary,

$\Rightarrow X_2' =X_2(\frac{N_1}{N_2})^2=0.05\times (\frac{500}{200})^2=0.05\times\frac{25}{4}=0.3125\Omega$

a) Equivalent resistance and reactance referred to as primary,

$R_{eq1}=R_1+R_2'=(0.3+0.1248)\Omega =0.4248\Omega$

$X_{eq1}=X_1+X_2'=2\Omega+0.3125\Omega =2.3125\Omega$

b) Equivalent resistance and reactance referred to as secondary coil,

$R_{eq2}=R_2+R_1'=(0.02+0.048)\Omega =0.068\Omega$

$X_{eq2}=X_2+X_1'=0.05\Omega+0.32\Omega =0.37\Omega$

c) Equivalent impedance referred to as primary side,

$z=\sqrt{R_{eq1}^2+X_{eq1}^2}=\sqrt{0.4248^2+2.3125^2}=2.35\Omega$

d) We know that output power factor = true power/apparent power

true power$(P)=\frac{E^2}{R_{eq2}}$

Apparent power $(P_a)=\frac{E^2}{Z}$

Output power factor $\frac{R_{eq2}}{Z}=\frac{0.068}{2.35}=0.0289$

$\Rightarrow \cos\phi=0.0289$

$\Rightarrow \phi=\cos^{-1}(0.0289)=88.34^\circ$

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, $S_{in}=I^2Z=2.35I^2$

Apparent output power $S_{out}=I^2Z_{o}$