Hi,

I need to find the Norton current of the attached circuit.

Here's what I've done:

First, I transformed the voltage source into a 9 A current source, with the 20 ohm resistor in parallel.
Using the current divider formula, we have 6 A going through that resistor.
Therefore, Isc = 3+3 = 6 A.

I input 6000 into Moodle (mA required) but it says the answer is wrong.
What am I doing wrong?

Thanks.

 

Because 6A is the wrong answer. Try to find thevenin's voltage (Vth) and Rth first.

 

So draw the transformed circuit, noting where the short is located so that you only include the current that actually goes through the short. You will probably see your mistake in short order, but if not then post your transformed circuit here so that we can help you identify the problem.

 

Hi,

I need to find the Norton current of the attached circuit.

Here's what I've done:

First, I transformed the voltage source into a 9 A current source, with the 20 ohm resistor in parallel.
Using the current divider formula, we have 6 A going through that resistor.
Therefore, Isc = 3+3 = 6 A.

I input 6000 into Moodle (mA required) but it says the answer is wrong.
What am I doing wrong?

Thanks.

Why are you only dividing the 9 A? If 9 A flows into that node from the 9 A source and splits into the 20 Ω and the 40 Ω paths, then were does the current flowing into that node from the 3 A source go?

Also, be careful about the polarity. When you say Isc is so much, is that flowing from 'a' to 'b' or from 'b' to 'a'?