Transformer MCQs Part 11

Contents

- 1 A 40 Ω resistive load is to be impedance matched by transformer to a source with internal resistance of 4000 Ω . If open circuit voltage of source is 60 V. Then the secondary power in watts will be
- 2 A low voltage electrical system employs a 300V to 75 step down transformer for safety. The equivalent resistance of electrical system is 15 ohms. Considering the transformer to be ideal and overall system as lossless and power system as 1, what is the power used
- 3 The no-load power factor of transformer is small due to which of following reason
- 4 With the increase in load, the eddy current
- 5 The transformer that has no electric isolation

## A 40 Ω resistive load is to be impedance matched by transformer to a source with internal resistance of 4000 Ω . If open circuit voltage of source is 60 V. Then the secondary power in watts will be

A 40 Ω is to be impedance matched by transformer to a source with internal resistance of 4000. If open circuit voltage of source is 60 V. Then the secondary power in watts will be:

- 0.02
- 0.075
- 0.095
- 0.2

Correct answer: 2. 0.075

Solution:

For maximum power transformer, half of source voltage is dropped in internal resistance of source.

Voltage across primary = 60/2 = 30

——————————————–

Zp = (N1/N2)^2 * Zs

4000 = (N1/N2)^2 * 40

N1/N2 = 10

———————————————

From Formula Vp/Vs = N1/N2

Vs = Vp/(N1/N2)

Vs = 30/10=3

Secondary power (Ps) = (Vs^2)/R = 3/40 = 0.075

## A low voltage electrical system employs a 300V to 75 step down transformer for safety. The equivalent resistance of electrical system is 15 ohms. Considering the transformer to be ideal and overall system as lossless and power system as 1, what is the power used

A low voltage electrical system employs a 300V to 75 step down transformer for safety. The equivalent resistance of electrical system is 15 ohms. Considering the transformer to be ideal and overall system as lossless and power system as 1, what is the power used.

- 200 W
- 375 W
- 600 W
- 1285 W

Correct answer: 2. 375 W

Explanation:

The above numerical can be solved using two methods. Either on parimary side or on secondary.

Method 1:

Secondary current (Is) = Vs/R = 75/15 = 5 A

P = Vs * Is * PF = 75 * 5 * 1 = 375 W

Method 2

Primary current (Ip) = (Vs * Is)/Vp = 1.25 A

P = Vp * Ip * PF = 300 * 1.25 * 1 = 375 W

## The no-load power factor of transformer is small due to which of following reason

The no-load power factor of transformer is small due to which of following reason:

- Magnetizing component of Io is large
- Iron loss component of Io is large
- Magnetizing component of Io is small
- None of these

Correct answer: 1. Magnetizing component of Io is large

## With the increase in load, the eddy current

With the increase in load, the eddy current

- Increases
- Decreases
- Remains unchanged
- None of these

Correct answer: 3. Remains unchanged

## The transformer that has no electric isolation

The transformer that has no electric isolation:

- Potential transformer
- Current transformer
- Power transformer
- Autotransformer

Correct answer: 4. Autotransformer

Explanation: The autotransformer is an electrical transformer with only one winding. In an autotransformer, portions of the same winding act as both the primary winding and secondary winding sides of the transformer, due to which there is no electric isolation between primary and secondary circuits.