# Voltage drop polarity

Consider the below circuit :

According to my textbook Vs will be

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?

• 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
– G36
Apr 21, 2019 at 12:27