# How to solve for V using source transformation method

Completely new to electric circuits, what is the simplest and beginner-friendly way to solve for V using source transformation?

I know that the 12 A current source could be transformed into voltage source, so:

12 * 8 = 96 V voltage source takes the place of the former 12 A current source, and the 8 Ω resistor is now in-series with the circuit.

I'm confused on how to solve for 7 A current source using source transformation method, what would the circuit even look like?

• Sorry if this looks like a dumb question, I may be overthinking it Apr 16 at 8:07

I'm confused on how to solve for 7 A current source using source transformation method, what would the circuit even look like?

It's the reverse process of what you did earlier. This time you determine the open circuit voltage of the 7 amp source and 14 Ω resistor (98 volts) and, put that 98 volts in series with 14 Ω: -

what is the simplest and beginner-friendly way to solve for V using source transformation?

Start with the 7 amp source and 14 Ω transformation and you end up with two resistors in series (14 Ω and 6 Ω). Then convert back to a current source in parallel with 20 Ω (4.9 amps): -

Now, you should be able to see that both current sources are in parallel and can be combined into one source of 16.9 amps. Can you take it from here?

• Hi, so continuing your explanation, I decide to not combine the current sources and transform both into voltage sources, leaving me with a circuit that's in-series (2 voltage sources Vs1 = 96 V, and Vs2 = 98 V. After doing mesh analysis I found the value of i = 0.071 A. and V = 0.071 * 8 = 0.568 V. Is this correct? Apr 16 at 11:00
• No that doesn't sound right. Far too small for the voltage. You should not transform the 8 ohm resistor if you are trying to find the voltage across it. The transformed voltage is not equivalent. Apr 16 at 11:14
• I'm sorry but I haven't been taught about combining parallel current sources, could you help on elaborating more on how to solve this? Apr 16 at 11:21
• I'd say so @anawss. Apr 16 at 13:42
• Thank you for your patience, Mr. Andy. Wish you all the best in life. Apr 16 at 15:13

The wanted voltage is between nodes A and Ground. The writer of the problem has intentionally drawn the wanted voltage like it somehow was a private property of a single resistor to confuse those who do not always remember that voltage is between 2 nodes.

What to do if it must be solved by using source transformations?

Let nodes A and Ground and the 6 ohm resistor stay where they are. Convert the current sources and their parallel resistors to voltage sources which have series resistors. Then you'll have a single loop circuit. Calculate the loop current with Ohm's Law. Calculate the voltage between A and Ground with Kirchhoff's voltage law.