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Here I began with Vth making mesh equations :

48-6I1-12(I1+I2)=0

12(I1+I2)+42+8I2=0

With which I get I1=4A & I2=-2A

Then I considered the open circuit part AB to use the 2 mesh currents there which is where I do not understand how to proceed.

For Rth also I am getting an error as well.

enter image description here

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  • \$\begingroup\$ I get Rth = 6.667 ohms. What did you get and what was the correct answer listed as? I get Vth(AB) = 32 volts. \$\endgroup\$ – Andy aka Dec 9 '18 at 18:07
  • \$\begingroup\$ Rth given is 4ohms I got about 5.333 though and yeah Vth is correct can you explain how you got the Vth \$\endgroup\$ – The Idiot Dec 9 '18 at 18:11
  • \$\begingroup\$ Yeah, sorry, my mistake Rth is 4 ohms. You don't really need to make mesh equations to solve this. \$\endgroup\$ – Andy aka Dec 9 '18 at 18:13
  • \$\begingroup\$ Could you explain it \$\endgroup\$ – The Idiot Dec 9 '18 at 18:16
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Start with this: -

enter image description here

The impedance into the node marked with a red arrow is \$\dfrac{1}{\frac{1}{12}+\frac{1}{6}}\$ = 4 ohms.

And this is in series with another 4 ohms when you look at node B. That makes 8 ohms but, there is a resistor to node C\$^1\$ of 8 ohms. So the total impedance between B and A is: -

\$\dfrac{1}{\frac{1}{8}+\frac{1}{8}}\$ = 4 ohms.


\$^1\$ node C connects to node A when analysing impedances because voltage sources short to zero volts. Can you get the voltage next yourself?

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  • \$\begingroup\$ yeah actually I got to the Rth part myself later on still stuck with the voltage though getting it as 28V somehow \$\endgroup\$ – The Idiot Dec 9 '18 at 18:26
  • \$\begingroup\$ Work on the voltage at the junction of the three restores where my arrow points. \$\endgroup\$ – Andy aka Dec 9 '18 at 21:07

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