# Sign actual current with mesh current for ac circuits

Take a look at the following circuit, using mesh method, we can have the following two equations \begin{align} (Z_1+Z_3) I_1 - Z_3I_2 &= V_1 \tag{1} \\ -Z_3 I_1 + (Z_2+Z_3)I_2 &= -V_2 \tag{2} \end{align} where $$\ Z_1=8,Z_2=j4,Z_3=-j2,V_1=40,V_2=j20 \$$. Solving the two equations, I get $$\ I_1=3+j4,I_2=-13-j4\$$. Now how can I know that the mesh currents in the same/opposite directions of the actual currents that pass $$\ V_1,V_2\$$? For this circuit, we can see the actual currents direction by looking at the polarities of the sources but I'm asking in general. I need this for computing the power at each source. In pure resistors circuits, obviously when the current is negative, it is opposite in direction with the actual current. The problem in ac circuits, current/voltage are complex numbers.

• Doesn't matter...You can assume any direction and in the end up with a -ve sign for the computed current if your assumed direction was wrong. Jul 9, 2021 at 4:55
• @MituRaj, not sure if you've read my post properly. Jul 9, 2021 at 6:47
• I did. You assume it to be true only for 'Pure resistor' circuits. But it's true if it involves linear elements like capacitor and inductors as well. Jul 9, 2021 at 7:03

Look at the following diagram:

simulate this circuit – Schematic created using CircuitLab

Above, I've labeled the polarities on the parts. The way I did this is to just "follow the current direction." The tail of the current is (+) and the head is (-). One way I remember reading about is that since a current through a dissipating device goes from the more positive end to the more negative end," you label the more positive end with (+), etc. If you examine the $$\I_1\$$ loop, you'll see that I followed this convention from the perspective of $$\I_1\$$. If you examine the $$\I_2\$$ loop, you'll see that I still followed this convention from the perspective of $$\I_2\$$, even when the signs disagree with the prior assignment made for $$\I_1\$$ ($$\C_1\$$ is the example, here.)

You found that $$\I_1=3+4j\$$ and that $$\I_2=-13-4j\$$. I agree with these.

It is easy to label two of the nodes, but the node ($$\V_3\$$) shared by $$\R_1\$$, $$\L_1\$$, and $$\C_1\$$ has to be computed. You can start from either end and work towards the middle.

For example, $$\V_3=V_1-I_1\cdot R_1=40\:\text{V} -\left(3+4j\right)\:\text{A} \cdot 8\:\Omega=\left(16-32j\right)\:\text{V}\$$.

Or, working it the other direction and noting the direction of the arrow on the diagram which is in the opposite direction so we must add the product term instead of subtracting it, find $$\V_3=V_2+I_2\cdot L_1={+}20j\:\text{V} +\left(-13-4j\right)\:\text{A} \cdot 4j\:\Omega=\left(16-32j\right)\:\text{V}\$$.

Now for power.

I guess a general procedure is to use the voltage difference times the complex conjugate of the current for each component in the circuit. The voltage difference is found by subtracting the node voltage at the tip of the current arrow from the node voltage at the tail of the current arrow.

• $$\V_1\$$: $$\P_{V_1}=\Delta V_1\cdot I_1^{^*}=\left(0\:\text{V}-40\:\text{V}\right)\cdot \left(3-4j\right)\:\text{A}={-}120+160j\$$.
• $$\V_2\$$: $$\P_{V_2}=\Delta V_2\cdot I_2^{^*}=\left(20j\:\text{V}-0\:\text{V}\right)\cdot \left(-13+4j\right)\:\text{A}={-}80-260j\$$.
• $$\R_1\$$: $$\P_{R_1}=\Delta V_{R_1}\cdot I_1^{^*}=\left(40\:\text{V}-\left(16-32j\right)\:\text{V}\right)\cdot \left(3-4j\right)\:\text{A}=200+0j\$$.
• $$\L_1\$$: $$\P_{L_1}=\Delta V_{L_1}\cdot I_2^{^*}=\left(\left(16-32j\right)\:\text{V}-20j\:\text{V}\right)\cdot \left(-13+4j\right)\:\text{A}=0+740j\$$.
• $$\C_1\$$: $$\P_{C_1}=\Delta V_{C_1}\cdot \left(I_1-I_2\right)^{^*}=\left(\left(16-32j\right)\:\text{V}-0\:\text{V}\right)\cdot \left(16-8j\right):\text{A}=0-640j\$$.

Tellegen's theorem says that the sum should always be zero:

\begin{align*} {-}120&+160j \\ {-}80&-260j \\ 200&+0j \\ 0&+740j \\ 0&-640j \\\hline 0&+0j \end{align*}

And so it is.

For a two-terminal device the rule for computing the power absorbed by the device is always:

simulate this circuit – Schematic created using CircuitLab

It turns out that for passive components (resistors, capacitors, inductors) the direction that you choose for the current arrow doesn't matter. That is, the formula you use won't change if you decide solve the problem with current arrow in the opposite direction.

However, for voltage sources you may have to use the negative of the voltage depending on which direction the current arrow is pointing relative to the voltage polarity. For instance, both of the following describe exactly the same situation but in the second case the A and B points are flipped because the current orientation has changed:

simulate this circuit

Note that we are calculating the power absorbed by the voltage source.

• Not sure if you're addressing my problem. I'm not having a problem with power formula, the problem with detecting whether the source is absorbing or supplying energy based on the direction of current. This is obviously based on the sign convention. In my example, both sources are supplying power to the circuit (i.e. negative values). Jul 9, 2021 at 6:41

You already have the current coming out of the positive terminal of the source on the left-hand side, that is your $$\I_1\$$.

The current coming out of the positive terminal of your source on the right hand side is the negative of your $$\I_2\$$, thus $$\13+j4\$$.

You can just assume an element of your circuit is supplying power. The math will show you if it is absorbing from the circuit or providing to the circuit.

In the figure below you assume voltage, $$\V\$$, with polarity as shown and current, $$\I\$$, with direction as shown. Then calculate complex power $$\S=VI^*=P+jQ\$$ from the actual values you found in your analysis. As you see in Jonk's answer, your source may be providing or absorbing real power ($$\P\$$) and may be providing or absorbing reactive power ($$\Q\$$) from the circuit.

• In my example, it is clear but I'm asking in general. For example, if we have multiple sources and impedances in series. Jul 9, 2021 at 3:35
• Hi @CroCo, it doesn't matter how complex the circuit. The only complicating factor is the complex math - but same exact principles. I've edited my answer to point out that in a circuit with complex elements ($X_C, X_L$) you use conjugate of current when calculating complex power $S$. This same $S=VI^*$ holds true for a purely resistive circuit too, it just simplifies to $VI$ since complex term is zero and conjugation does nothing to the real part. Jul 9, 2021 at 12:12