How does source transformation convert a voltage source into an equivalent current source?

In one line

A practical voltage source \( V \) with series resistance \( R \) can be replaced by a current source \( I = V/R \) with the same \( R \) in parallel, and vice versa, without changing anything for the load.

Explained simply

Think of two different chargers that look identical from the outside: plug any phone into either and it gets exactly the same voltage and current. The load cannot tell which one is inside the box. Source transformation swaps one "box" for the other to make the circuit easier to solve.

Technically, two sources are equivalent if they give the same terminal voltage and current to every load. A voltage source \( V \) in series with \( R \) and a current source \( I = V/R \) in parallel with \( R \) both have open-circuit voltage \( V \) and short-circuit current \( V/R \), so they are equivalent with respect to the terminals.

How it works

1. Voltage to current: \( I = V/R \); keep \( R \) but connect it in parallel. The current arrow points toward the terminal that was the + terminal of the voltage source.
2. Current to voltage: \( V = IR \); keep \( R \) but connect it in series. The + terminal is on the side the current arrow points to.
3. Use it repeatedly to combine sources: parallel current sources add, series voltage sources add, and resistors combine as usual.
4. Only practical sources (with a resistance) can be transformed. An ideal source alone has no equivalent of the other kind.

Key formula

\( I = \dfrac{V}{R} \quad\text{or}\quad V = IR \)

\( V \) = source voltage (V), \( I \) = source current (A), \( R \) = the same resistance (\( \Omega \)) in both forms.

Quick example

10 V in series with 2 \( \Omega \) becomes 5 A in parallel with 2 \( \Omega \). Check with a 3 \( \Omega \) load. Voltage form: \( I_L = 10/(2+3) = 2 \) A. Current form: current divider \( I_L = 5 \times 2/(2+3) = 2 \) A. Same answer, so the forms are equivalent.

Common mistakes

• Changing the value of \( R \) during the transformation; it stays the same.
• Getting the current direction wrong: it must push current out of the original + terminal.
• Transforming the resistor whose current or voltage is asked; you then lose the quantity you wanted.

How MU asks it

Asked as a numerical part (about 5–8 marks): "Using source transformation, find the current through the 5 \( \Omega \) resistor." For full marks draw a fresh circuit after every transformation step, write \( I = V/R \) values each time, and finish with a single loop or current divider. Also asked as a short theory question with a diagram of both forms.