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The problem * source transformation is just a combining and rearranging voltage or current sources. I don't think its a universal term.

My approach so far:

and I'm stuck. any hint?

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Now solving the circuit using mesh analysis:-

For Left Loop:-

160 Im +200 Im+120 (Im-Is) = 24 .....eq (1)

For Right Loop:-

120 (Is-Im) + 190 Is = 36 + Vs .....eq (2)


=>

Since Is = 0.25A & putting in eq(1)
Im= 9/80 = 112.5 mA
=> Vs = 28 V

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    \$\begingroup\$ I'm marking this as the correct answer. The other answers are also helpful, but this one is very comprehensive. Thank you. \$\endgroup\$ – taiduckman Sep 18 '12 at 23:03
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You are forgetting to "connect" the resistor back once you're transforming your sources, that's the problem.

V IN SERIES with R = Is PARALLEL with R

For example, on the right part of your circuit, you can swap the 0.2A source and the 160Ω resistor with a Vs=(0.2×160) = 32V source AND the same RESISTOR, IN SERIES with that voltage source, and vice versa. So, you would have the 32V IN SERIES with -8 = 24V.

That part is right. But, you would have 160Ω in parallel with 200Ω (that's the part you are doing wrong). You're just forgetting to connect back the resistor once you transform your sources.

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