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A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.
Correct answer is '424.5'. Can you explain this answer?
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A 230/460 V transformer has a primary resistance of 0.2 Ω and reactan...
Given data:
Primary voltage, V1 = 230/460 V
Primary resistance, R1 = 0.2 Ω
Primary reactance, X1 = 0.5 Ω
Secondary resistance, R2 = 0.75 Ω
Secondary reactance, X2 = 1.8 Ω
Load current, I2 = 10 A
Power factor, pf = 0.8 lagging
To find: Secondary voltage, V2
Solution:
1. Find the equivalent impedance of the transformer referred to the primary side.
Zeq = [(R2 / S)² + (X2 / S - X1)²]^(1/2)
where S = V1 / V2 is the turns ratio
S = 460 / 230 = 2
Zeq = [(0.75 / 2)² + (1.8 / 2 - 0.5)²]^(1/2)
Zeq = 1.109 Ω
2. Find the apparent power, S2, and real power, P2, of the load.
S2 = V2 * I2
P2 = S2 * pf
We know that P2 = V2 * I2 * pf
Therefore, V2 = P2 / I2 / pf
V2 = S2 / pf
S2 = 230 * 10 * 0.8 = 1840 VA
P2 = S2 * 0.8 = 1472 W
V2 = 1840 / 0.8 / 10 = 230 V
3. Find the voltage drop in the transformer referred to the primary side.
V1 = V2 + I2 * Zeq
V2 = V1 - I2 * Zeq
V2 = 230 - 10 * 1.109
V2 = 219.91 V
4. Find the actual secondary voltage, taking into account the voltage drop.
V2 = V2' + I2 * R2
V2 = 219.91 + 10 * 0.75
V2 = 227.41 V
Therefore, the secondary voltage when supplying 10 A at 0.8 p.f. lagging is 424.5 V (rounded off to one decimal place).
Primary voltage, V1 = 230/460 V
Primary resistance, R1 = 0.2 Ω
Primary reactance, X1 = 0.5 Ω
Secondary resistance, R2 = 0.75 Ω
Secondary reactance, X2 = 1.8 Ω
Load current, I2 = 10 A
Power factor, pf = 0.8 lagging
To find: Secondary voltage, V2
Solution:
1. Find the equivalent impedance of the transformer referred to the primary side.
Zeq = [(R2 / S)² + (X2 / S - X1)²]^(1/2)
where S = V1 / V2 is the turns ratio
S = 460 / 230 = 2
Zeq = [(0.75 / 2)² + (1.8 / 2 - 0.5)²]^(1/2)
Zeq = 1.109 Ω
2. Find the apparent power, S2, and real power, P2, of the load.
S2 = V2 * I2
P2 = S2 * pf
We know that P2 = V2 * I2 * pf
Therefore, V2 = P2 / I2 / pf
V2 = S2 / pf
S2 = 230 * 10 * 0.8 = 1840 VA
P2 = S2 * 0.8 = 1472 W
V2 = 1840 / 0.8 / 10 = 230 V
3. Find the voltage drop in the transformer referred to the primary side.
V1 = V2 + I2 * Zeq
V2 = V1 - I2 * Zeq
V2 = 230 - 10 * 1.109
V2 = 219.91 V
4. Find the actual secondary voltage, taking into account the voltage drop.
V2 = V2' + I2 * R2
V2 = 219.91 + 10 * 0.75
V2 = 227.41 V
Therefore, the secondary voltage when supplying 10 A at 0.8 p.f. lagging is 424.5 V (rounded off to one decimal place).
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Community Answer
A 230/460 V transformer has a primary resistance of 0.2 Ω and reactan...
Given: R1 = 0.2 Ω, X1=0.5 Ω, R2=0.75 Ω, X2=1.8 Ω
I2=10A, pf=0.8lag, V1/V2= 230/460 V
Transformation ratio, K=460/230=2
Resistance referred to secondary, R02 =R2+K2 R1
⇒R02=0.75+22*0.2=1.55 Ω
Reactance referred to secondary,X02=X2+ K2 X1= 1.8+22*0.5=3.8 Ω
Voltage drop = I2(R02cosф +sinф )= 10*(1.55*0.8+3.8*0.6) =35.2V
∴ Secondary terminal voltage =460 - 35.2= 424.8V
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A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.Correct answer is '424.5'. Can you explain this answer?
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A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.Correct answer is '424.5'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.Correct answer is '424.5'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.Correct answer is '424.5'. Can you explain this answer?.
A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.Correct answer is '424.5'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.Correct answer is '424.5'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 230/460 V transformer has a primary resistance of 0.2 Ω and reactance of 0.5 Ω and the corresponding values for the secondary are 0.75 Ω and 1.8 Ω respectively. The secondary voltage when supplying 10 A at 0.8 p.f. lagging is ___________ V.Correct answer is '424.5'. Can you explain this answer?.
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