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a 5 kva, 200/1000 v, 50 hz, single-phase transformer gave the following test results: o.c. test (lv side): 2000 v, 1.2 a, 90 w s.c. test (hv side) : 50v, 5 a, 110 w (i) calculate the parameters of the equivalent circuit referred to the lv side (ii) calculate the output secondary voltage when delivering 3 kw at 0.8 pf lagging, the input primary voltage being 200 v. find the percentage regulation also

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Answer:

(i):

R_eq ≈ 2777.78 ohms

X_eq = 10 ohms

X_m ≈ 1666.67 ohms

(ii):

N = 1/5

3000 VA = 200 V * (5 * I_secondary)

V_secondary ≈ -8133.34 V - j30 V

Final:

The percentage regulation is approximately 100%

Step-by-step explanation:

To find the parameters of the equivalent circuit referred to the low-voltage (LV) side, you can use the open-circuit (O.C.) and short-circuit (S.C.) test results.

(i) Parameters of the Equivalent Circuit Referred to the LV Side:

1. Equivalent Resistance (R_eq):

Use the O.C. test results:

Voltage (V_oc) = 2000 V

Current (I_oc) = 1.2 A

Power (P_oc) = 90 W

R_eq = (V_oc / I_oc)^2 / P_oc

R_eq = (2000 V / 1.2 A)^2 / 90 W

R_eq ≈ 2777.78 ohms

2. Equivalent Reactance (X_eq):

Use the S.C. test results:

Voltage (V_sc) = 50 V

Current (I_sc) = 5 A

Power (P_sc) = 110 W

X_eq = V_sc / I_sc

X_eq = 50 V / 5 A

X_eq = 10 ohms

3. Magnetizing Reactance (X_m):

X_m can be found using the O.C. test results:

X_m = V_oc / I_oc

X_m = 2000 V / 1.2 A

X_m ≈ 1666.67 ohms

(ii) To calculate the output secondary voltage when delivering 3 kW at 0.8 power factor (lagging), you can use the following formula for the apparent power (S):

S = V_primary * I_primary = V_secondary * I_secondary

Given:

- Input primary voltage (V_primary) = 200 V

- Output power (P_out) = 3 kW (3000 VA)

- Power factor (PF) = 0.8 (lagging)

First, calculate the output current (I_secondary):

P_out = V_secondary * I_secondary * PF

I_secondary = P_out / (V_secondary * PF)

I_secondary = 3000 VA / (V_secondary * 0.8)

Next, calculate the current on the primary side (I_primary) using the turns ratio (N):

N = V_primary / V_secondary

I_primary = I_secondary / N

Now, calculate N:

N = V_primary / V_secondary

N = 200 V / 1000 V

N = 1/5

I_primary = I_secondary / (1/5)

I_primary = 5 * I_secondary

Now, you can calculate I_primary:

I_primary = 5 * I_secondary

Substitute this into the equation for apparent power (S):

S = V_primary * I_primary

3000 VA = 200 V * (5 * I_secondary)

Solve for I_secondary:

I_secondary = 3000 VA / (200 V * 5) = 3 A

Now that you have I_secondary, you can find the secondary voltage (V_secondary) using the formula:

V_secondary = V_primary - (I_secondary * (R_eq + jX_eq))

V_secondary = 200 V - (3 A * (2777.78 ohms + j10 ohms))

V_secondary ≈ 200 V - (8333.34 ohms + j30 ohms)

V_secondary ≈ 200 V - (8333.34 ohms + j30 ohms)

V_secondary ≈ 200 V - 8333.34 ohms - j30 ohms

V_secondary ≈ (200 V - 8333.34 ohms) - j30 ohms

V_secondary ≈ -8133.34 V - j30 V

Now, to find the percentage regulation, use the formula:

Percentage Regulation = ((V_no_load - V_full_load) / V_full_load) * 100%

V_no_load is the no-load secondary voltage, and V_full_load is the calculated secondary voltage when delivering 3 kW.

Percentage Regulation = ((-8133.34 V - 0 V) / -8133.34 V) * 100%

Percentage Regulation ≈ (8133.34 V / 8133.34 V) * 100%

Percentage Regulation ≈ 100%

The percentage regulation is approximately 100%.

User PVitt
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