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Source Transformations - GATE EE Network Theory Free MCQ Test with solutions


MCQ Practice Test & Solutions: Test: Source Transformations (10 Questions)

You can prepare effectively for Electrical Engineering (EE) Network Theory (Electric Circuits) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Source Transformations". These 10 questions have been designed by the experts with the latest curriculum of Electrical Engineering (EE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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Test: Source Transformations - Question 1

A voltage source connected in series with a resistor can be converted to a?

Detailed Solution: Question 1

A voltage source connected in series can be converted to a current source connected in parallel using the relation obtained from ohm’s law, that is V=IR. This equation shows that a voltage source connected in series has the same impact as a current source connected in parallel.

Test: Source Transformations - Question 2

Calculate the total current in the circuit.

Detailed Solution: Question 2

The 9mA source connected in parallel to the 5 kohm resistor can be converted to a 45V source in series with a 5 kohm resistor. Applying mesh analysis, we get:
I=(45-3)/(5+4.7+3)= 3.3mA.

Test: Source Transformations - Question 3

Find the value of voltage once source transformation is applied to the circuit.​

Detailed Solution: Question 3

Using ohm’s law, we can use the relation: V=IR.
Thus V=10*5= 50V.

Test: Source Transformations - Question 4

What will the value of the current be once source transformation is applied to the circuit?

Detailed Solution: Question 4

Using ohm’s law, we can use the relation: V=IR.
Thus I=V/R.
I=220/22=10A.

Test: Source Transformations - Question 5

Once the circuit is transformed to a voltage source where will the resistance be connected?


 

Detailed Solution: Question 5

The resistance is connected in series with the voltage source because we are transforming a current source connected in parallel to a resistor to a voltage source connected in series with it.

Test: Source Transformations - Question 6

A current source connected in parallel with a resistor can be converted to a?

Detailed Solution: Question 6

 A current source connected in parallel can be converted to a voltage source connected in series using the relation obtained from ohm’s law, that is V=IR. This equation shows that a current source connected in parallel has the same impact as a voltage source connected in series.

Test: Source Transformations - Question 7

 A source transformation is_________

Detailed Solution: Question 7

A source transformation is bilateral because a voltage source can be converted to a current source and vice-versa.

Test: Source Transformations - Question 8

 In source transformation________

Detailed Solution: Question 8

 In source transformation, the value of the voltage and current sources change when changed from voltage to current source and current to voltage source but the value of the resistances remains the same.

Test: Source Transformations - Question 9

 If there are 3 10V sources connected in parallel then on source transformation__________

Detailed Solution: Question 9

When voltages are connected in parallel, the effect of only one source is considered because the effect of the voltage remains the same when connected in parallel.

Test: Source Transformations - Question 10

The current through a 2 Ω resister will be

Detailed Solution: Question 10

Concept of Series Source:

When an ideal voltage source and an ideal current source in series, the combination has an ideal current source property.

The current in the circuit is independent of any element connected in series to it.

In a series circuit, the current flows through all the elements is the same. Thus, any element connected in series with an ideal current source is redundant and it is equivalent to an ideal current source only.

Application:

The given circuit can be drawn as,

Since Here 4 V voltage source and 5 A current source are in series hence, We have to consider current source only,

Now circuit becomes,

From the circuit, it is clear that the current through 2 Ω resistance will be 5 A.

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