Find the Thevenin Equivalent circuit

[Fatima Hasan]

I don't know where that value of 2.173 came from, so I can't explain. You'll have to tell me how you calculated I2.

R4 and R3 in parallel --> R* = 2*10/(2+10) = 5/3Ω
R* and R2 in series --> R total = 5/3 +5 = 20/3Ω
I1 = R total / (R total + R 1)* i_s
I1 = (20/3) / ((20/3)+5) * 5 = 2.857 A
I_s = I1 + I2
I2 = 5 - 2.857 = 2.143 A

 

[The Electrician]

We got the value of 6.25 volts across R2 with RL removed. If RL is not there then R3 and R4 are not in parallel; R3 and R4 are only in parallel if RL is replaced with a short. You are trying to compare the voltage across R2 with RL removed to the voltage across R2 with RL replaced with a short. You can't expect the voltage in the two cases to be the same.

 

[Fatima Hasan]

We got the value of 6.25 volts across R2 with RL removed. If RL is not there then R3 and R4 are not in parallel; R3 and R4 are only in parallel if RL is replaced with a short. You are trying to compare the voltage across R2 with RL removed to the voltage across R2 with RL replaced with a short. You can't expect the voltage in the two cases to be the same.

OK , Thank you for your help

 

[gneill]

Since the OP has reached a correct result I can offer an alternative method to get to the solution.

You can use successive source transformations (essentially Norton to Thevenin or Thevenin to Norton transformations), working your way from left to right through the circuit while "absorbing" new resistances along the way. For example, you can convert the original current source and its parallel resistor to a voltage source with a series resistor:

As you can see, the "new" 5 Ω resistor can be combined with R2 into a single 10 Ω resistor and the result is a 25 V voltage source in series with 10 Ω. That pair can in turn be converted to its Norton equivalent of a current source in parallel with a resistor and you'll then find that its resistor will be in parallel with R3, so you can combine them. Continue in such a fashion until you've absorbed all the components into a single Thevenin or Norton model driving the load.

 

[Fatima Hasan]

Since the OP has reached a correct result I can offer an alternative method to get to the solution.

You can use successive source transformations (essentially Norton to Thevenin or Thevenin to Norton transformations), working your way from left to right through the circuit while "absorbing" new resistances along the way. For example, you can convert the original current source and its parallel resistor to a voltage source with a series resistor:

View attachment 222127

As you can see, the "new" 5 Ω resistor can be combined with R2 into a single 10 Ω resistor and the result is a 25 V voltage source in series with 10 Ω. That pair can in turn be converted to its Norton equivalent of a current source in parallel with a resistor and you'll then find that its resistor will be in parallel with R3, so you can combine them. Continue in such a fashion until you've absorbed all the components into a single Thevenin or Norton model driving the load.

Thanks for the explanation.
Can I know what software did you use to draw the circuits?

 

[gneill]

Thanks for the explanation.
Can I know what software did you use to draw the circuits?

I used an ancient copy of Microsoft Visio. Still works great for quickly putting together a drawing to accompany a physics problem.