# Kirchhoff’s Laws

## Electric Potential in Circuits

Electric potential roughly represents the concentration of electrical potential energy in a circuit. Potential is also similar to water pressure.

We call a change in electric potential across a circuit element the electric potential difference or more commonly the voltage.

Electric potential quickly spreads to a uniform value throughout an uninterrupted section of wire. This is similar to how water in a cup stays at the same height because it spreads out against the force of gravity.

Electric potential is constant until it reaches a circuit element.
Across a resistor the potential drops, so the voltage is negative.
Across a battery the potential increases, so the voltage is positive.

Example: In the diagram above the potential in red is 4V. The potential in grey is 2V. What is the potential difference across the resistor?
solution $$\Delta V = V_f-V_i$$ $$\Delta V = 2V-4V$$ $$\Delta V = -2V$$

What is the voltage across the resistor?
solution

Voltage means a potential difference. They are the same.

$$\Delta V = V_f-V_i$$ $$\Delta V = 2V-4V$$ $$\Delta V = -2V$$
Example: Calculate the resistance using the diagram above.
solution $$\Delta V = V_f-V_i$$ $$\Delta V = 0.70V-1.50V$$ $$\Delta V = -0.80V$$
$$\Delta V=IR$$ $$R = \frac{\Delta V}{I}$$ $$R = \frac{0.80}{0.0020}$$ $$R = 400 \Omega$$
Example: When a wire branches, the potential is the same for an uninterrupted section. Use the potential to find the potential difference, or voltage drop, across each resistor.
solution $$\Delta V = V_f-V_i$$ $$\Delta V = 5.5V-9V$$ The current is to the left, because the potential is dropping from right to left and resistors always decrease voltage.

Investigation: Explain why voltages in parallel, like in the diagram above, must all be the same?
solution
Electric potential is the same for all of an uninterrupted wire, even if it branches. So, each branch has the same potential. This means they must also have the same potential difference.
Example: The electric potential for each uninterrupted section of wire is shown in the circuit diagram above. Use the potential to find the potential difference, or voltage, across each element in the circuit.
solution

Subtract the potential before and after each element to find potential difference across each element.

$$V_f-V_i = \Delta V$$ $$2V-6V = -4V$$ $$0V-2V = -2V$$ $$6V-0V = 6V$$

## Kirchhoff’s Law: Voltage

If it is possible for the current to take a looping path through the circuit, the total change in potential is zero. Otherwise the potential would keep getting higher.

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 $$\sum V = 0$$ $$V_{bat}+V_1+V_2+V_3 = 0$$ $$V_{bat} = -V_1-V_2-V_3$$ $$V_{bat} = 1.164+3.88+0.97$$ $$V_{4} = 6.014V$$
Example: Find the missing resistance.
solution

Only one of the 3.4V drops are in each loop.

$$\sum V = 0$$ $$9 - 3.4 - V = 0$$ $$V = 5.6 V$$
$$V=IR$$ $$R = \frac{V}{I}$$ $$R = \frac{5.6}{0.0005}$$ $$R=11, \! 200 \Omega$$

## Kirchhoff’s Law: Current

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

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

$$I_{in}$$ = total charge entering a point per second [A, amps]
$$I_{out}$$ = total charge exiting a point per second [A, amps]

If a circuit doesn't branch it will have the same current at every point.

Example: If a 3-way circuit junction has two wires with 2A each entering. How much current is in the 3rd wire?
solution $$\sum I _{in} = \sum I_{out}$$ $$2A + 2A = I_{out}$$ $$4A = I_{out}$$
Example: The current is 10 mA before two 166 Ω resistors. What is the current after the resistors?
solution
The total current entering a part of a circuit must equal the total current exiting. If a circuit doesn't branch it will have the same current at every point. Current doesn't change across resistors and batteries. $$I = 10 \, mA$$
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

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

## 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]

Wiring more resistors in parallel results in lower equivalent resistance.

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 $$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$ $$\frac{1}{10} = \frac{1}{30} + \frac{1}{R_{2}}$$ $$\frac{1}{10} - \frac{1}{30} = \frac{1}{R_{2}}$$ $$\frac{3}{30} - \frac{1}{30} = \frac{1}{R_{2}}$$ $$\frac{2}{30} = \frac{1}{R_{2}}$$ $$15k \Omega = R_{2}$$
Example: Identify groups of resistors that are in series or parallel. Combine them to shrink the circuit until you reduce the circuit to just one resistor.
solution

Equivalent Resistance = 266.6Ω

## 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. Let's 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.

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

Looks good!