Consider the below circuit :

enter image description here

According to my textbook Vs will be

enter image description here

However I have another solution which is different from this.

Using superposition, we will first find the drop due to the voltage source and due to its polarity, Vs will have a positive terminal on the bottom of the resistor and negative terminal on top of the resistor.

So due to this source we will have a drop of (50/50+150) * (j26.6m.) For the current source, due to its direction, also the positive terminal of Vs will still be at the bottom and the negative terminal will be at the top, so the drop will have the same sign and will be equal to (50*150/50+150) *(j0.1168m) and the total drop due to these sources will be +j11.03mv not with a negative sign as the answer.

Can someone explain why my answer is different from that of the solution?

  • \$\begingroup\$ The Vs voltage due to Voltage source will be negative and the same is true for a current source. Hence Vs = - ((50/(50+150)) x j26.6m) + (- (50*150)/(50+150)) x j0.1168m) = - j11.03mV \$\endgroup\$
    – G36
    Apr 21, 2019 at 12:27

1 Answer 1


The negative sign in the answer indicates that the polarity of the voltage is opposite to that marked on the diagram.

You mention that twice in your working, but forgot to include it in the result.

Vs will have a positive terminal on the bottom of the resistor and negative terminal on top of the resistor.

After completing the calculations an noting the polarity is opposite that shown, you should invert the number you got to get the final result

  • \$\begingroup\$ I know that , but this means that the answer should always have a negative sign ? Or it just depends on the convention we use to indicate the polarity of the voltage drop Vs ? I mean can we assume that Vs has polarity opposite to that in the diagram and in this case the answer will be correct ? \$\endgroup\$
    – John adams
    Apr 21, 2019 at 10:24
  • 2
    \$\begingroup\$ I think the convention used in Identifying Vs is as if we measure the voltage drop with a multi-meter we put the positive wire in the top and the negative wire on the bottom of the resistor but we will have a negative sign on the multi-meter because the polarity should be opposite due to the induced sources. Is this right ? \$\endgroup\$
    – John adams
    Apr 21, 2019 at 10:24
  • \$\begingroup\$ yes. that's right. a negative values is the opposite of a positive value \$\endgroup\$ Apr 21, 2019 at 10:45

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