# Kirchhoff’s Laws

Gustav Kirchhoff
Gustav Kirchhoff was the physicist that is given the credit for creating the “Kirchhoff’s Law.” He was born on May 12, 1824 in Konigsberg, Prussia (now Russia). This law is an extension of Ohm’s Law and generalized the equations regarding current flow and adapted them in the case of electrical conductors in three dimensions. Eventually, Kirchhoff found that current flows through a conductor at the speed of light. These laws are extremely important in understanding how an electric circuit functions.

• Kirchhoff also contributed to the study of spectroscopy and was a major contributor to the advancement in the research of blackbody radiation.
• Kirchhoff also founded the spectrum analysis with chemist Robert Bunsen
• “Kirchhoff’s Law” was initially called “Laws of Closed Electric Circuits”, but was later named after the creator.
• Kirchhoff was first a “Privatdozent”, which is an unpaid lecturer at the University of Berlin. But he eventually got a position at the University of Heidelberg as the professor of Physics.

• -Ariana Dideban

## Kirchhoff’s Law: Current

For any point on a circuit: the total charge flowing into the point equals the total charge flowing out.

# $$I _{in} = I_{out}$$

$$I_{in}$$ = total charge entering a circuit per second [A]
$$I_{out}$$ = total charge exiting a circuit per second [A]
Example: If a 3-way circuit junction has two wires with 2A each entering. How much current is in the 3rd wire?
solution $$I _{in} = I_{out}$$ $$2A + 2A = I_{out}$$ $$4A = I_{out}$$
Example: A circuit splits into 3 branches. Branch #1 has 20mA, branch #2 has 40mA, and branch #3 has 55mA. Find the current before the split.
solution

no need to convert units for just addition and subtraction

$$I _{in} = I_{out}$$ $$I _{in} = 20 + 40 + 55$$ $$I _{in} = 115 mA$$
Example: Find the current in the top branch.
solution $$I _{in} = I_{out}$$ $$450 = I_{1} + 115 + 120$$ $$450 - 115 - 120 = I_{1}$$ $$215 mA = I_{1}$$

## Kirchhoff’s Law: Voltage

DC voltage sources, like batteries, provide positive voltage. Most other elements have negative voltage. The elements with negative voltage convert the electric potential into some other type of energy, like heat, light, sound, or motion.

Check here to see what the different circuit diagram symbols mean.
(mostly resistors or DC voltage sources)

For any closed circuit loop the sum of all potential differences equals zero.

# $$\sum V = 0$$ $$V_1+V_2+V_3+V_4 = 0$$

$$V$$ = potential difference, voltage [V, volts]
Example: What's the voltage drop across the resistor in the diagram?
solution $$V_1+V_2+V_3+V_4 = 0$$ $$1.51-0.55-0.33-V_{4} = 0$$ $$V_{4} = -0.63V$$
Example: Use Ohm's law to find the voltage for each resistor
solution

All elements in a series have the same current

$$V = IR$$ $$V = (0.000194)(6000)$$ $$V = 1.164V$$
$$V = IR$$ $$V = (0.000194)(20000)$$ $$V = 3.88V$$
$$V = IR$$ $$V = (0.000194)(5000)$$ $$V = 0.97V$$

Use the resistor voltages to calculate the voltage in the battery.
solution $$\sum V = 0$$ $$V_{bat}+V_1+V_2+V_3 = 0$$ $$V_{bat} = -V_1-V_2-V_3$$ $$V_{bat} = 1.164+3.88+0.97$$ $$V_{4} = 6.014V$$

## Resistors in Series

Electrical components are in a series when they are connected in a single path so that all charge flows through the same components.

The total equivalent resistance for resistors in a series is the sum of the resistors. This means you can replace several resistors in a series with just one and the circuit will be the same.

# $$R_{eq} = R_1+R_2+ ... + R_n$$

$$R_{n}$$ = A single resistor in a series [ohms, Ω]
$$R_{eq}$$ = Equivalent resistance. The resistance of a single resistor that could replace several resistors. [ohms, Ω]
Example: What is the equivalent resistance for four 100Ω resistors in a series?
solution $$R_{eq} = R_1+R_2+R_3+R_4$$ $$R_{eq} = 100+100+100+100$$ $$R_{eq} = 400\Omega$$

## Resistors in Parallel

Electrical components are in parallel when the path branches, and charges take different paths.

The equation for replacing resistors in parallel is a bit more complex. The inverse of the total equivalent resistance for resistors in parallel is equal to the sum of the inverse of each resistance.

# $$\frac{1}{R_{eq}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+ ... + \frac{1}{R_{n}}$$

$$R_{n}$$ = A single resistor in parallel [ohms, Ω]
$$R_{eq}$$ = Equivalent resistance. The resistance of a single resistor that could replace several resistors [ohms, Ω]
Example: What is the equivalent resistance for five 100Ω resistors all in parallel?
solution $$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} + \frac{1}{R_{5}}$$ $$\frac{1}{R_{eq}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} + \frac{1}{100} + \frac{1}{100}$$ $$\frac{1}{R_{eq}} = \frac{5}{100}$$ $$R_{eq} = \frac{100}{5}$$ $$R_{eq} = 20 \Omega$$
Example: Find the equivalent resistance for the three resistors above.
solution $$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$$ $$\frac{1}{R_{eq}} = \frac{1}{400} + \frac{1}{200} + \frac{1}{100}$$ $$\frac{1}{R_{eq}} = \frac{1}{400} + \left(\frac{2}{2}\right)\frac{1}{200} + \left(\frac{4}{4}\right)\frac{1}{100}$$ $$\frac{1}{R_{eq}} = \frac{1}{400} + \frac{2}{400} + \frac{4}{400}$$ $$\frac{1}{R_{eq}} = \frac{7}{400}$$ $$R_{eq} = \frac{400}{7}$$ $$R_{eq} = 57.14 \Omega$$
Example: The equivalent resistance for the above circuit is 10kΩ. Use that information to find the missing resistance.
solution $$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$ $$\frac{1}{10} = \frac{1}{30} + \frac{1}{R_{2}}$$ $$\frac{1}{10} - \frac{1}{30} = \frac{1}{R_{2}}$$ $$\frac{3}{30} - \frac{1}{30} = \frac{1}{R_{2}}$$ $$\frac{2}{30} = \frac{1}{R_{2}}$$ $$15k \Omega = R_{2}$$

## Solving Complex Circuits

How do you solve a circuit with both series and parallel elements? One technique is to simplify the circuit by replacing resistors in series or parallel with one equivalent resistor.

We can start by combining the two parallel resistors.

$$\frac{1}{R_{eq}} = \frac{1}{R_{2}} + \frac{1}{R_{3}}$$ $$R_{eq} = \left(\frac{1}{800} + \frac{1}{900}\right)^{-1}$$ $$R_{eq} = \left(0.00125 + 0.00\overline{11}\right)^{-1}$$ $$R_{eq} = 420 \Omega$$

Next we can combine the 3 resistors in series.

$$R_{eq} = 600+420+500$$ $$R_{eq} = 1520 \Omega$$ Kirchoff's Voltage law says that the total positive voltage must equal the negative voltage. This tells us the voltage drop on the resistor is the same as the battery.

$$V_{bat} + V_1 = 0$$ $$6.0 + V_1 = 0$$ $$V_1 = -6.0$$

We can solve for current through the resistor with Ohm's law.

$$V=IR$$ $$I = \frac{V}{R}$$ $$I = \frac{6.0}{1530}$$ $$I = 0.0039A$$

This current can be applied to any circuit element in series with Req. If we expand the circuit back to when they were all in a series we will know the current for all the resistors.

We can use Ohm's law to find voltages.

$$V=IR_1$$ $$V=(0.0039)(600)$$ $$V=2.3V$$
$$V=IR_4$$ $$V=(0.0039)(500)$$ $$V=2.0V$$
$$V=IR_{eq}$$ $$V=(0.0039)(420)$$ $$V=1.7V$$

Resistors in parallel have the same voltage, but not the same current. Lets expand back to our full size circuit and fill in the voltage.

We can use Ohm's law on R2 and R3 to find current.

$$I=\frac{V}{R_2}$$ $$I=\frac{1.7}{800}$$ $$I=0.0021A$$
$$I=\frac{V}{R_3}$$ $$I=\frac{1.7}{900}$$ $$I=0.0019A$$

We can calculate the power dissipated by each element with P=IV.

$$P = IV_1$$ $$P = (0.0039)(2.3)$$ $$P = 0.0090W$$
$$P = IV_4$$ $$P = (0.0039)(2.0)$$ $$P = 0.0078W$$
$$P = IV_2$$ $$P = (0.0021)(1.7)$$ $$P = 0.0036W$$
$$P = IV_3$$ $$P = (0.0019)(1.7)$$ $$P = 0.0032W$$

That was a long problem. Let's check our work. The total voltage should equal zero, if we count the resistors in parallel as one.

$$V_1+V_4+V_{eq} + V_{bat} = 0$$ $$-2.3-2.0-1.7+6.0 = 0$$ $$0 = 0$$

The power from the battery should equal the sum of the power lost in the resistors.

$$P_{bat}=IV$$ $$P_{bat}=(0.0039)(6.0)$$ $$P_{bat}=0.023W$$
$$P_{resistors} = 0.0090W+0.0078W+0.0036W+0.0032W$$ $$P_{resistors} = 0.023W$$

Looks good!