# How is a capacitor implemented in Kirchoff's Loop Rule?

I'm real stumped here.

I may be missing something, but I'm genuinely perplexed as to how one would add a capacitor to the list of elements being added to an equation through Kirchoff's Loop Rule (KVL).

Suppose a 10V power supply (E1) with an internal resistance of, say, 3 ohms is connected to a resistor (R1) of any resistance, and this resistor is connected to a capacitor (C1) of 24uF in a closed loop - how would KVL be used here? I understand that the KVL equation begins with E1-R1, but how does the capacitor fit in? I can't seem to find a good explanation.

If someone could walk me through it, I'd greatly appreciate it. Thanks in advance!

• Kirchohff's loop rule says that in a closed loop, the sum of voltage differences across the circuit elements is zero. In a capacitor the voltage difference is given as $V = Q/C$. You can add it like this in the sum! – WarreG Jan 10 at 13:09
• "power supply (E1) with an internal resistance of, say, 3V" Why is a resistance given in volts? – Steeven Jan 10 at 14:05
• Oh! My bad, I'll change that now. – K. Millar Jan 11 at 16:36

Kirchhoff's voltage law (or loop law) is simply that the sum of all voltages around a loop must be zero:

$$\sum v=0$$

In more intuitive terms, all "used voltage" must be "provided", for example by a power supply, and all "provided voltage" must also be "used up", otherwise charges would constantly accelerate somewhere.

We can think of "used voltage" as negative and "provided voltage" as positive. Or we can, to be more accurate, move around the loop, say, clockwise and give opposite signs to opposite polarities (if a voltage is seen as "plus to minus" then positive and if as "minus to plus" then negative, as we move through the loop). Then simply add them all up according to the law:

$$\sum v=0\quad\Leftrightarrow\\-v_{\text{resistor1}}-v_{\text{resistor2}}-v_{\text{resistor3}}+v_{\text{power supply}}-v_{\text{capacitor1}}-v_{\text{capacitor2}}\,\cdots=0$$

As this example equation shows, naturally all resistors will "use" voltage (there is a voltage drop across them), so negative. All power supplies such as batteries etc. "provide" voltage, so positive. And capacitors likewise often "use" voltage, so negative (but keep an eye on them in each case to be sure).

Now, simply plug in formulas for each term in this equation:

• Resistances that follow Ohm's law can be replaced with $$v=Ri$$,
• the voltage of power supplies is often given, like in your case $$v=10\;\mathrm V$$,
• and as @WarreG suggests in a comment above, capacitors likewise often have plug-and-play formulas, such as $$v=Q/C$$.

How you pick a formula depends on which parameters you know already. When the equation is filled in, you are done and can solve for unknowns or whichever parameter you are looking for in this equation.

You will need to specify what kind of voltage is involved. The following assumes a 10 vrms sinusoidal voltage. You will also need to know the frequency, since the impedance of a capacitor depends on frequency.

The resistance (R) of a resistor is a special case of electrical impedance (Z). The current through and voltage across a resistor are said to be “in phase”. The impedance of a resistor is simply its magnitude in ohms. That is

$$Z_{R}=R$$

Capacitors and inductors are circuit elements in which the current and voltage are 90 degrees out of phase. The magnitude of the impedance of a capacitor is called its capacitive reactance, $$X_C$$ and given by

$$X_{C}=\frac{1}{2πfC}$$

Where $$f$$ is the frequency of the voltage and C the capacitance.

Its impedance is given by

$$Z_{C}=-jX_{C}=\frac{-j}{2πfC}$$

$$j$$ is the imaginary number equal to the square root of minus 1. The j accounts for the fact that the impedance is -90 degrees in the complex plane. The impedance of a resistor is a real number in the complex plane.

When applying KVL to your series RC circuit, in phasor notation, you will have

$$+10-i3-iR_{1}-(-jX_{C})i=0$$

Where $$i$$ is the sinusoidal current, and it is assumed your 10 volt power supply is a 10 vrms sinusoidal voltage source with no phase angle (v=0 when t=0). To find the current, divide the voltage by the equivalent series impedance, $$Z_{equiv}$$

$$Z_{equiv} = (R_{1} + 3)-jX_{C})$$

The rest is complex number algebra.

Hope this helps.