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I'm doing some homework sheets, and I just wanted to know if the given answer to this problems is wrong:

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I'm getting V0 (the node) as 144v and the current ia as 48A, I just wanted to confirm if the error is in the question, or if it was me who went wrong:

Here is my working out: $$ \frac{6I_a-V_0}{3} = I_a $$ $$ I_a = \frac{V_0}{3} $$ Then doing KCL at node V0, treating the node below the 24ohm resistor as ground, and subbing in Ia gives:

$$ \frac{V_0-60}{2} + \frac{V_0}{24} + \frac{V_0-6I_a}{3} = 0 $$ $$ \frac{V_0}{2} - 30 + \frac{V_0}{24} + \frac{V_0}{3} - \frac{2V_0}{3} = 0 $$ $$ \frac{5V_0}{24} = 30 $$ And then then gives V0 as 144V and Ia as 48A.

Thanks guys!

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  • \$\begingroup\$ Please note that checking whether the solution is correct does not require solving the circuit. Just use the provided solution and check if V=IR is satisfied at every resistor. It is relatively quick to see that voltages of 60V, 48V, 36V and currents of 6A, 2A, (-)4A, 12A are self-consistent. \$\endgroup\$ – Ben Voigt Jan 21 '17 at 15:27
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V is ground referenced, 6ia is not. Therefore 6ia-V/3 = ia is wrong. Try again. This is a classic trap in these questions. Floating voltage sources must be dealt with care.

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  • \$\begingroup\$ This assumes (as you have also) that one end of 24Ohm R is deaignated as ground. \$\endgroup\$ – Adil Malik Jan 21 '17 at 15:17
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Your first equation makes your ground at the positive terminal of 60V or negative terminal of the current dependent voltage source. From there, your KCL at Vo is wrong since, as you said, your ground is the bottom node.

You could adjust either of the equations just as long as you follow your designated ground. It would be helpful to choose and label your ground before proceeding with nodal analysis so you won't get confused.

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