How is phase angle calculated in an RLC circuit?

If I have a source with a phase angle, like this how may I count the phase angle of the voltages and currents across all the components, as well as the returning current? What is the angle between voltage and current at each component, as well as between source voltage and returning current?

• Which components!? Schematic or it didn't happen! – stevenvh May 11 '12 at 15:26

First year electronics?

Draw out your circuit.

Give each components an impedance coeficient X1,X2,X3 etc.
As you write out the value for each component add a multiplier to convert it to the 's' domain (Laplace) For Capacitors use '1/s', for Inductors use 's' for Resistors use '1'. Be consistent with the direction of current, and decide to start if currents produce negative or positive voltages (it does not matter which, so long as you are consistent)

Write a set of simultanious equations for each loop in the circuit. Imagine current flowing into each node. Use a rule that sum total of voltage in each loop is zero, the sum total of current in each loop is zero and the fact that V=I*X (where x is an impedance)

Now you need to solve those equations to get an equation for each of the nodal voltages and currents in each wire.

The above results in linear algebra, which you can solve as a system of equations or using matrix solving. I use a PC tool such as MathCAD when things get too complicated (which they can rather easily)

As an optional step, convert your equations back into the time domain using inverse Laplace.

Finally as you have an equation for voltage and each current you can apply an input and calculate the current and voltage in each component complete with phase.

Here is a worked answer that makes use of the fact that the question is simplified, no need for mesh currents etc.

From inspection you can see four current, I1, I2, I3 and I4 in V, R, C and L respecitvely.
We can easily write these currents:

I2=-V2/R

I3=-V2/XC
-> I3=-V2/(1/(s*C))
-> I3=-C*V2*s

I4=-V2/XL
-> I4=-V2/(s*L)

I1=I2+I3+I4

V2=10V@30deg convert to a complex representation:
V2=10V(sin(30deg) + j*cos(30deg)
V2=8.66+5j (V)

Now calculate I2, I3, I4, the currents in the components, e.g.

We can use the fact that s = j*w (j * ohmega)
and w = 2 * pi * f

I3=-1e-6 * (8.66+5j) * (0 + j(2 * 3.14159 * 300))
I3=0.425e-3 - 0.016j (A)
|I3|=0.019A
atan2(-0.016,0.425e-3)=150deg
so I3 in the capacitor is 0.019A @ 150deg

The voltage across the capacitor is the same as the supply

The same can be done for I2 and I4.
I1=I2+I3+I4 and so is what the ammeter at U4 is measuring

• So there's no "direct" way to account for phase shift imposed by the various components? That is to say, I cannot simply say +90 for L and -90 for C? – Joe Stavitsky May 11 '12 at 17:15
• Ahh, now you have given a schematic, it makes things easier. The answer is simple because the question is simple, see my updated answer. – Jason Morgan May 11 '12 at 18:17

First calculate impedance of circuit in frequency domain . $$L \rightarrow LS \space or \space jLw$$ $$c \rightarrow \frac{1}{CS} \space or \space \frac{1}{jwC}$$ $$Impedance \space of \space circuit \space Z(jw)=|Z(jw)|\angle Z(jw)$$ $$\angle Z(jw)$$ angle of impedance is angle between voltage and current . $$Z=\frac{|V|\angle V}{|I| \angle I}=\left|\frac{V}{I}\right|\angle(\angle V -\angle I)$$