Test: Source Transformations - Electrical Engineering (EE) MCQ

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.

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.

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

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

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.

 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.

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

 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.

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.

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.