Kirchhoff’s Laws were published in 1845 by German physicist Gustav Kirchhoff. When Kirchhoff's laws are combined with Ohm's law we are able to calculate voltage and current for complex circuits.

Electric Potential in Circuits

Electric potential roughly represents the concentration of energy in a circuit. The 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.

Differences in electric potential are called voltage.

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 4 V. The potential in grey is 2 V. What is the potential difference across the resistor?

solution

$$\Delta V = V_f-V_i $$ $$\Delta V = 2\, \mathrm{V}-4\, \mathrm{V}$$ $$\Delta V = -2\, \mathrm{V}$$

What is the voltage across the resistor?

solution

Voltage means a potential difference. They are the same.

$$\Delta V = V_f-V_i $$ $$\Delta V = 2\, \mathrm{V}-4\, \mathrm{V}$$ $$\Delta V = -2\, \mathrm{V}$$

Example: What voltage is the battery providing?

solution

The potential jumps from 0 to 1.5 V. The battery adds 1.5 volts to the circuit.

Example: Use Ohm's law to calculate the resistance in the diagram above.

solution

$$\Delta V = V_f-V_i $$ $$\Delta V = 0.70\, \mathrm{V}-1.50\, \mathrm{V}$$ $$\Delta V = -0.80\, \mathrm{V}$$
$$\Delta V=IR$$ $$R = \frac{\Delta V}{I}$$ $$R = \frac{0.80}{0.0020}$$ $$R = 400 \, \Omega$$

Example: When a wire branches, the potential is the same for an uninterrupted section. Use the potential to find the potential difference, or voltage, across each resistor.

solution

$$\Delta V = V_f-V_i $$ $$\Delta V = 5.5\, \mathrm{V}-9\, \mathrm{V}$$

The current is to the left, because the potential is dropping from right to left and resistors always decrease voltage.

Question: What direction is the current?

answer

The current is directed to the left. Current is the flow of charge, and charge flows from high potential to low potential.

Question: Which path has the highest current? Why?

answer

The 100 Ω resistor has the highest current

All three paths have the same voltage, so the only difference is the resistance. Resistance makes it harder for current to flow. The lowest resistor will have the highest current.

Example: Calculate the current across each resistor.

solution

$$\Delta V = IR$$ $$I = \frac{\Delta V}{R}$$ $$I = \frac{1.5 \, \mathrm{V}}{400 \, \Omega} \quad \enspace \quad I = \frac{1.5 \, \mathrm{V}}{200 \, \Omega} \quad \enspace \quad I = \frac{1.5 \, \mathrm{V}}{100 \, \Omega}$$ $$I = 0.00375 \, \mathrm{A} \quad \quad I = 0.0075 \, \mathrm{A} \quad \quad I = 0.015 \, \mathrm{A} $$ $$I = 3.75 \, \mathrm{mA} \quad \quad I = 7.5 \, \mathrm{mA} \quad \quad I = 15 \, \mathrm{mA} $$

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.

$$V_f-V_i = \Delta V $$ $$2\, \mathrm{V}-6\, \mathrm{V} = -4\, \mathrm{V}$$ $$0\, \mathrm{V}-2\, \mathrm{V} = -2\, \mathrm{V}$$ $$6\, \mathrm{V}-0\, \mathrm{V} = 6\, \mathrm{V}$$

You might have noticed that the total voltage for either path the current could take adds up to zero. This is an important principle.

$$+6 \, \mathrm{V} -4 \, \mathrm{V} -2 \, \mathrm{V} = 0$$

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 voltages equals zero.

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

$V$ = potential difference, voltage [V, volts]

Kirchhoff's Voltage law is a consequence of conservation of energy. Voltage is electric potential energy per charge. As current flows through the circuit, total energy doesn't change.

Example: What's the voltage drop across the resistor in the diagram?

hint

The total voltage for any looping path must add up to zero.

$$ V_1+V_2+V_3+V_4 = 0 $$

solution

$$ V_1+V_2+V_3+V_4 = 0 $$ $$ 1.51-0.55-0.33-V_{4} = 0 $$ $$ V_{4} = -0.63\, \mathrm{V} $$

Example: Use Ohm's law to find the voltage for each resistor.

solution

All elements in a series have the same current

$$V = IR$$ $$V = (0.000194)(6000)$$ $$V = 1.164\, \mathrm{V}$$
$$V = IR$$ $$V = (0.000194)(20000)$$ $$V = 3.88\, \mathrm{V}$$
$$V = IR$$ $$V = (0.000194)(5000)$$ $$V = 0.97\, \mathrm{V}$$

Use the resistor voltages to calculate the voltage in the battery.

solution

$$ \sum V = 0$$ $$ V_{bat}+V_1+V_2+V_3 = 0 $$ $$ V_{bat} = -V_1-V_2-V_3 $$ $$ V_{bat} = 1.164+3.88+0.97 $$ $$ V_{4} = 6.014\, \mathrm{V} $$

Example: Find the missing resistance.

solution

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

$$ \sum V = 0$$ $$ 9 - 3.4 - V = 0 $$ $$V = 5.6 \, \mathrm{V}$$
$$V=IR$$ $$R = \frac{V}{I}$$ $$R = \frac{5.6}{0.0005}$$ $$R=11200 \, \Omega$$

Kirchhoff’s Law: Current

Kirchhoff’s current law is true because charge is conserved. Total charge can't increase or decrease.

At any point on a circuit, the total charge flowing in equals the total charge flowing out.

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

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

Sections of a circuit that don't branch will have the same current everywhere.

Example: If a 3-way circuit junction has two wires with 2 A each entering. How much current is in the 3rd wire?

solution

$$ I _{\text{in}} = I_{\text{out}} $$ $$ 2\, \mathrm{A} + 2\, \mathrm{A} = I_{\text{out}} $$ $$ 4\, \mathrm{A} = I_{\text{out}} $$

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.

Example: Find the current before the circuit splits into 3 branches.

solution

no need to convert units for just addition and subtraction

$$ I _{\text{in}} = I_{\text{out}} $$ $$ I _{\text{in}} = 20\, \mathrm{mA} + 40\, \mathrm{mA} + 55\, \mathrm{mA} $$ $$ I _{\text{in}} = 115 \, \mathrm{mA} $$

Example: Find the current in the top branch.

solution

$$ I _{\text{in}} = I_{\text{out}} $$ $$ 450\, \mathrm{mA} = I_{1} + 115\, \mathrm{mA} + 120\, \mathrm{mA} $$ $$450\, \mathrm{mA} - 115\, \mathrm{mA} - 120\, \mathrm{mA} = I_{1}$$ $$ 215 \, \mathrm{mA} = I_{1}$$

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.

$$R_{eq} = R_1+R_2+R_3+\cdots$$

$R_{n}$ = 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]

Adding resistors in series increases the total resistance.

Example: You can replace resistors in a series with one equivalent resistor. What one resistor could replace these four resistors?

solution

$$R_{eq} = R_1+R_2+R_3+R_4$$ $$R_{eq} = 120\, \Omega + 150\, \Omega + 200\, \Omega + 100\, \Omega$$ $$R_{eq} = 570\, \Omega$$

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.

$$\frac{1}{R_{eq}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+ \frac{1}{R_{3}}+\cdots$$

$R_{n}$ = A single resistor in parallel [ohms, Ω]
$R_{eq}$ = Equivalent resistance. The resistance of a single resistor that could replace several resistors [ohms, Ω]

Adding resistors in parallel lowers the total resistance. This makes sense if you think of each parallel resistor as a possible path for current. More paths allow more current and less total resistance.

Example: What is the equivalent resistance for five 100 Ω resistors all in parallel with each other?

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 10 kΩ. Use that information to find the missing resistance.

solution

$$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$ $$\frac{1}{10} = \frac{1}{30} + \frac{1}{R_{2}}$$ $$\frac{1}{10} - \frac{1}{30} = \frac{1}{R_{2}}$$ $$\frac{3}{30} - \frac{1}{30} = \frac{1}{R_{2}}$$ $$\frac{2}{30} = \frac{1}{R_{2}}$$ $$15\, \mathrm{k} \Omega = R_{2}$$

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.
(Start with the highlighted resistors in series.)

solution

$$ \text{Equivalent Resistance} = 266.6 \, \Omega $$

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.

finding resistance for a simplified circuit

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.

$$R_{eq} = 600+420+500$$ $$R_{eq} = 1520 \, \Omega$$

using Kirchoff's laws to build back up to the full circuit

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.

$$V_{bat} + V_1 = 0$$ $$6.0 \, \mathrm{V}+ V_1 = 0$$ $$V_1 = -6.0 \, \mathrm{V}$$

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

$$V=IR$$ $$I = \frac{V}{R}$$ $$I = \frac{6.0\, \mathrm{V}}{1520\, \Omega}$$ $$I = 0.0039\, \mathrm{A}$$

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.

$$V=IR_1$$ $$V=(0.0039\, \mathrm{A})(600\, \Omega)$$ $$V=2.3\, \mathrm{V}$$

$$V=IR_4$$ $$V=(0.0039\, \mathrm{A})(500\, \Omega)$$ $$V=2.0\, \mathrm{V}$$

$$V=IR_{eq}$$ $$V=(0.0039\, \mathrm{A})(420\, \Omega)$$ $$V=1.7\, \mathrm{V}$$

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.

$$I=\frac{V}{R_2}$$ $$I=\frac{1.7}{800}$$ $$I=0.0021 \, \mathrm{A}$$

$$I=\frac{V}{R_3}$$ $$I=\frac{1.7}{900}$$ $$I=0.0019 \, \mathrm{A}$$

solving for power dissipated

We can calculate the power dissipated by each element.

$$ \quad P = IV$$ $$P = (0.0039)(2.3)$$ $$P = 0.0090\, \mathrm{W}$$

$$\text{\color{#f06}{(R4)}} \quad P = IV$$ $$P = (0.0039)(2.0)$$ $$P = 0.0078\, \mathrm{W}$$

$$\text{\color{#f06}{(R2)}} \quad P = IV$$ $$P = (0.0021)(1.7)$$ $$P = 0.0036\, \mathrm{W}$$

$$\text{\color{#f06}{(R3)}} \quad P = IV$$ $$P = (0.0019)(1.7)$$ $$P = 0.0032\, \mathrm{W}$$

checking our work

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.

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

Looks good!

Example: Find the voltage of the battery.

strategy

Replace the three resistors in parallel with one equivalent resistor. The equivalent resistor will have the same voltage as each resistor in parallel because of Kirchhoff's voltage law.

Use Ohm's law to find the current through the equivalent resistor. That will be the same current that flows through the 30 kΩ resistor. We will then be able to find the voltage in that resistor with Ohm's law.

The battery's voltage is equal to the sum of the voltages across each resistor because of Kirchoff's voltage law.

solution

$$ \frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$$ $$ \frac{1}{R_{\mathrm{eq}}} = \frac{1}{80}+\frac{1}{90}+\frac{1}{90}$$ $$ R_{\mathrm{eq}} = {\color{#e30}28.8 \, \mathrm{k} \Omega}$$

Voltage is the same in parallel. The equivalent resistance has 3.0 V because the 80 kΩ resistor has 3.0 V.

$$V = I R_{\mathrm{eq}}$$ $$I= \frac{V}{R_{\mathrm{eq}}}$$ $$I= \frac{3.0 \, \mathrm{A}}{28\,800 \, \Omega}$$ $$I= 0.000104 \, \mathrm{A}= {\color{#08c}0.104 \, \mathrm{mA}}$$

The current in a series is the same, so we also know the current in the 30 kΩ resistor. With current and resistance we can calculate the voltage too.

$$V=IR$$ $$V=(0.000104 \, \mathrm{A})(30\,000 \, \Omega)$$ $$V= {\color{#0a7}3.125 \, \mathrm{V}}$$

We can use Kirchhoff’s voltage law to find the voltage in the battery.

$$0=V_{\mathrm{bat}} + V_1 + V_2$$ $$V_{\mathrm{bat}} = -V_1 - V_2$$ $$V_{\mathrm{bat}} = 3.0 \, \mathrm{V} + 3.125 \, \mathrm{V}$$ $$V_{\mathrm{bat}} = {\color{#c09} 6.125 \, \mathrm{V}} $$

practice problems (28)

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

Convert kΩ to Ω and mA to A before calculating.

printout.pdf show topics and mark harder problems

Example: A small sensor circuit has one side of a resistor at 9.0 V and the other side at 4.0 V. What is the potential difference across the resistor?

solution

Potential difference is the change in electric potential from one side to the other.

$$\Delta V = V_2 - V_1$$ $$\Delta V = 4.0 - 9.0$$ $$\Delta V = -5.0\,\mathrm{V}$$

The resistor has a 5.0 V drop.

Example: A battery raises charge from 0.0 V to 12.0 V. What potential difference does the battery provide?

solution

$$\Delta V = V_2 - V_1$$ $$\Delta V = 12.0 - 0.0$$ $$\Delta V = 12.0\,\mathrm{V}$$

The battery provides a 12.0 V potential rise.

Example: A loop has a 9.0 V battery and three voltage changes around the loop. Two resistor drops are -2.0 V and -3.5 V. What must the third voltage change be?

solution

Kirchhoff's voltage law says the voltage changes around a complete loop add to zero.

$$\sum V = 0$$ $$9.0 - 2.0 - 3.5 + V_3 = 0$$ $$3.5 + V_3 = 0$$ $$V_3 = -3.5\,\mathrm{V}$$

The missing voltage change is a 3.5 V drop.

Example: At a junction in a model train layout, 2.4 A flows in. Current leaves along three branches. Two branches carry 0.80 A and 0.60 A. How much current leaves in the third branch?

solution

Kirchhoff's current law says the current into a junction equals the current out of the junction.

$$I_{\mathrm{in}} = I_{\mathrm{out}}$$ $$2.4 = 0.80 + 0.60 + I_3$$ $$2.4 = 1.40 + I_3$$ $$I_3 = 1.0\,\mathrm{A}$$

The third branch carries 1.0 A.

Example: Two wires bring current into a junction. One wire brings in 0.50 A. Two wires leave the junction with currents of 1.2 A and 0.70 A. What current must come in through the second incoming wire?

solution

$$I_{\mathrm{in}} = I_{\mathrm{out}}$$ $$0.50 + I_2 = 1.2 + 0.70$$ $$0.50 + I_2 = 1.90$$ $$I_2 = 1.40\,\mathrm{A}$$

The second incoming wire carries 1.40 A.

Example: A circuit loop has a 9.0 V battery and two resistors labeled 100 Ω and 220 Ω. What is the current in the 220 Ω resistor?

solution

This cannot be solved from the information given. The circuit does not say whether the resistors are in series, in parallel, or arranged some other way.

Those different arrangements give different voltages across the 220 Ω resistor, so the current cannot be determined from the numbers alone.

Example: Three resistors are connected in series: 120 Ω, 330 Ω, and 470 Ω. What is the equivalent resistance?

solution

In series, resistances add directly.

$$R_{\mathrm{eq}} = R_1 + R_2 + R_3$$ $$R_{\mathrm{eq}} = 120 + 330 + 470$$ $$R_{\mathrm{eq}} = 920\,\Omega$$

Example: A repair kit has three resistors in series: 1.5 kΩ, 680 Ω, and 2.2 kΩ. What is the equivalent resistance in kΩ?

solution

Convert the kΩ values to Ω first.

$$1.5\,\mathrm{k}\Omega = 1500\,\Omega$$ $$2.2\,\mathrm{k}\Omega = 2200\,\Omega$$ $$R_{\mathrm{eq}} = 1500 + 680 + 2200$$ $$R_{\mathrm{eq}} = 4380$$ $$R_{\mathrm{eq}} = 4.38\,\mathrm{k}\Omega$$

Example: Two resistors, 300 Ω and 600 Ω, are connected in parallel. What is their equivalent resistance?

solution

$$\frac{1}{R_{\mathrm{eq}}} = \frac{1}{300} + \frac{1}{600}$$ $$\frac{1}{R_{\mathrm{eq}}} = 0.00333 + 0.00167$$ $$\frac{1}{R_{\mathrm{eq}}} = 0.00500$$ $$R_{\mathrm{eq}} = 200\,\Omega$$

Example: A dashboard light circuit uses a 4.0 kΩ resistor in parallel with a 12 kΩ resistor. What is the equivalent resistance?

solution

Convert to the same unit before calculating.

$$4.0\,\mathrm{k}\Omega = 4000\,\Omega$$ $$12\,\mathrm{k}\Omega = 12000\,\Omega$$ $$\frac{1}{R_{\mathrm{eq}}} = \frac{1}{4000} + \frac{1}{12000}$$ $$\frac{1}{R_{\mathrm{eq}}} = 0.000333$$ $$R_{\mathrm{eq}} = 3000$$ $$R_{\mathrm{eq}} = 3.0\,\mathrm{k}\Omega$$

Example: You need 120 Ω for a circuit, but you only have a 300 Ω resistor. What resistor should you put in parallel with it?

solution

$$\frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$\frac{1}{120} = \frac{1}{300} + \frac{1}{R_2}$$ $$\frac{1}{R_2} = \frac{1}{120} - \frac{1}{300}$$ $$\frac{1}{R_2} = 0.00500$$ $$R_2 = 200\,\Omega$$

Adding a resistor in parallel always lowers the equivalent resistance below the smallest resistor, so the answer has to make R_eq less than 300 Ω.

Example: A 12 V battery powers three series resistors: 100 Ω, 200 Ω, and 300 Ω. What is the circuit current, and what is the voltage across the 200 Ω resistor?

solution

$$R_{\mathrm{eq}} = 100 + 200 + 300$$ $$R_{\mathrm{eq}} = 600$$ $$I = \frac{V}{R}$$ $$I = \frac{12}{600}$$ $$I = 0.020\,\mathrm{A}$$ $$V_2 = IR_2$$ $$V_2 = (0.020)(200)$$ $$V_2 = 4.0\,\mathrm{V}$$

The current is 0.020 A, and the 200 Ω resistor has a 4.0 V drop.

Example: A 12 V power supply is connected across two parallel branches. One branch has a 2.0 kΩ resistor, and the other has a 3.0 kΩ resistor. What is the total current from the supply?

solution

$$2.0\,\mathrm{k}\Omega = 2000\,\Omega$$ $$3.0\,\mathrm{k}\Omega = 3000\,\Omega$$ $$I_1 = \frac{12}{2000}$$ $$I_1 = 0.0060$$ $$I_2 = \frac{12}{3000}$$ $$I_2 = 0.0040$$ $$I_t = 0.0060 + 0.0040$$ $$I_t = 0.010\,\mathrm{A}$$

The supply provides 0.010 A, or 10 mA.

Example: Two parallel branches are connected across a 9.0 V battery. One branch has a 1.0 kΩ resistor. The total current from the battery is 15 mA. What resistance is in the second branch?

solution

$$1.0\,\mathrm{k}\Omega = 1000\,\Omega$$ $$15\,\mathrm{mA} = 0.015\,\mathrm{A}$$ $$I_1 = \frac{9.0}{1000}$$ $$I_1 = 0.0090$$ $$I_2 = 0.015 - 0.0090$$ $$I_2 = 0.0060$$ $$R_2 = \frac{V}{I_2}$$ $$R_2 = \frac{9.0}{0.0060}$$ $$R_2 = 1500\,\Omega$$

The second branch has a resistance of 1.5 kΩ.

Example: A 10 V supply is connected to a 200 Ω resistor in series with a parallel pair of two 600 Ω resistors. What is the total current from the supply?

solution

First reduce the parallel pair.

$$\frac{1}{R_p} = \frac{1}{600} + \frac{1}{600}$$ $$\frac{1}{R_p} = 0.00333$$ $$R_p = 300$$

Now add the series resistor.

$$R_t = 200 + 300$$ $$R_t = 500$$ $$I = \frac{V}{R_t}$$ $$I = \frac{10}{500}$$ $$I = 0.020\,\mathrm{A}$$

Example: Find the equivalent resistance of the circuit above. Then find the total current from the 9.0 V battery.

solution

The 600 Ω and 1200 Ω resistors are in parallel.

$$\frac{1}{R_p} = \frac{1}{600} + \frac{1}{1200}$$ $$\frac{1}{R_p} = 0.00250$$ $$R_p = 400$$

That parallel section is in series with the 300 Ω and 200 Ω resistors.

$$R_t = 300 + 400 + 200$$ $$R_t = 900\,\Omega$$ $$I = \frac{V}{R_t}$$ $$I = \frac{9.0}{900}$$ $$I = 0.010\,\mathrm{A}$$

The total current is 0.010 A, or 10 mA.

Example: Find the voltage of the battery in the circuit above.

solution

The three vertical resistors are in parallel, so each has 4.0 V across it.

$$60\,\mathrm{k}\Omega = 60000\,\Omega$$ $$120\,\mathrm{k}\Omega = 120000\,\Omega$$ $$20\,\mathrm{k}\Omega = 20000\,\Omega$$ $$I_1 = \frac{4.0}{60000}$$ $$I_1 = 0.0000667$$ $$I_2 = \frac{4.0}{120000}$$ $$I_2 = 0.0000333$$ $$I_3 = \frac{4.0}{120000}$$ $$I_3 = 0.0000333$$ $$I_t = I_1 + I_2 + I_3$$ $$I_t = 0.000133$$

That total current flows through the 20 kΩ resistor.

$$V_4 = I_tR_4$$ $$V_4 = (0.000133)(20000)$$ $$V_4 = 2.67$$ $$V_b = 4.0 + 2.67$$ $$V_b = 6.67\,\mathrm{V}$$

Example: Find the missing resistance at the bottom of the circuit above.

solution

The parallel section has 2.4 V across each branch.

$$40\,\mathrm{k}\Omega = 40000\,\Omega$$ $$80\,\mathrm{k}\Omega = 80000\,\Omega$$ $$120\,\mathrm{k}\Omega = 120000\,\Omega$$ $$I_1 = \frac{2.4}{40000}$$ $$I_1 = 0.000060$$ $$I_2 = \frac{2.4}{80000}$$ $$I_2 = 0.000030$$ $$I_3 = \frac{2.4}{120000}$$ $$I_3 = 0.000020$$ $$I_t = I_1 + I_2 + I_3$$ $$I_t = 0.000110$$

The bottom resistor must use the rest of the battery voltage.

$$V_4 = 4.05 - 2.4$$ $$V_4 = 1.65$$ $$R_4 = \frac{V_4}{I_t}$$ $$R_4 = \frac{1.65}{0.000110}$$ $$R_4 = 15000$$ $$R_4 = 15\,\mathrm{k}\Omega$$

Example: Find the current from the battery in the circuit above.

solution

The parallel section has a 5.0 V drop. The rest of the battery voltage is across the 50 kΩ resistor at the bottom.

$$50\,\mathrm{k}\Omega = 50000\,\Omega$$ $$V_4 = 10.0 - 5.0$$ $$V_4 = 5.0$$ $$I = \frac{V_4}{R_4}$$ $$I = \frac{5.0}{50000}$$ $$I = 0.000100\,\mathrm{A}$$

The current from the battery is 0.100 mA.

Example: In the circuit above, R3 and R4 are in series with each other, and that pair is in parallel with R2. Find the current through R4 and the voltage across R4.

solution

R3 and R4 are on the same wire with no junction between them, so they are in series.

$$R_{34} = R_3 + R_4$$ $$R_{34} = 200 + 400$$ $$R_{34} = 600\,\Omega$$

That 600 Ω branch is in parallel with R2.

$$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_{34}}$$ $$\frac{1}{R_p} = \frac{1}{300} + \frac{1}{600}$$ $$\frac{1}{R_p} = 0.00500$$ $$R_p = 200\,\Omega$$

The parallel section is in series with R1.

$$R_{\mathrm{eq}} = R_1 + R_p$$ $$R_{\mathrm{eq}} = 100 + 200$$ $$R_{\mathrm{eq}} = 300\,\Omega$$

Use Ohm's law to find the current from the battery. All of it flows through R1.

$$I = \frac{V}{R_{\mathrm{eq}}}$$ $$I = \frac{9.0}{300}$$ $$I = 0.030\,\mathrm{A}$$ $$V_1 = IR_1$$ $$V_1 = (0.030)(100)$$ $$V_1 = 3.0\,\mathrm{V}$$

Kirchhoff's voltage law says the rest of the battery's 9.0 V is across the parallel section. Both branches get the same voltage.

$$V_p = 9.0 - 3.0$$ $$V_p = 6.0\,\mathrm{V}$$

The right branch has 6.0 V across its 600 Ω, so its current is:

$$I_{34} = \frac{V_p}{R_{34}}$$ $$I_{34} = \frac{6.0}{600}$$ $$I_{34} = 0.010\,\mathrm{A}$$

R3 and R4 are in series, so both carry 0.010 A.

$$V_4 = I_{34}R_4$$ $$V_4 = (0.010)(400)$$ $$V_4 = 4.0\,\mathrm{V}$$

Check with Kirchhoff's laws. R2 carries 6.0 V / 300 Ω = 0.020 A, and 0.020 A + 0.010 A = 0.030 A, which matches the battery current. Around the right loop, 9.0 − 3.0 − 2.0 − 4.0 = 0, since R3 drops 2.0 V.

Example: Find the current from the battery and the current through R4 in the circuit above.

solution

R1 connects straight across the battery, so it has the full 12 V across it.

$$I_1 = \frac{V}{R_1}$$ $$I_1 = \frac{12}{600}$$ $$I_1 = 0.020\,\mathrm{A}$$

The rest of the circuit is R2 in series with the parallel pair R3 and R4. Combine the pair first.

$$\frac{1}{R_{34}} = \frac{1}{300} + \frac{1}{600}$$ $$\frac{1}{R_{34}} = 0.00500$$ $$R_{34} = 200\,\Omega$$ $$R_{\mathrm{branch}} = R_2 + R_{34}$$ $$R_{\mathrm{branch}} = 200 + 200$$ $$R_{\mathrm{branch}} = 400\,\Omega$$

That branch also has the full 12 V across it.

$$I_{\mathrm{branch}} = \frac{12}{400}$$ $$I_{\mathrm{branch}} = 0.030\,\mathrm{A}$$

Kirchhoff's current law adds the two currents leaving the battery.

$$I = I_1 + I_{\mathrm{branch}}$$ $$I = 0.020 + 0.030$$ $$I = 0.050\,\mathrm{A}$$

To find the current in R4, find the voltage across the parallel pair.

$$V_2 = I_{\mathrm{branch}}R_2$$ $$V_2 = (0.030)(200)$$ $$V_2 = 6.0\,\mathrm{V}$$ $$V_{34} = 12 - 6.0$$ $$V_{34} = 6.0\,\mathrm{V}$$ $$I_4 = \frac{V_{34}}{R_4}$$ $$I_4 = \frac{6.0}{600}$$ $$I_4 = 0.010\,\mathrm{A}$$

Check: R3 carries 6.0 V / 300 Ω = 0.020 A, and 0.020 A + 0.010 A = 0.030 A, which matches the current through R2.

Example: An ammeter shows 10 mA through R2 in the circuit above. What is the battery voltage?

solution

R1 and R2 are in parallel, so they have the same voltage. Use R2's current to find it.

$$10\,\mathrm{mA} = 0.010\,\mathrm{A}$$
$$V_{12} = I_2R_2$$ $$V_{12} = (0.010)(600)$$ $$V_{12} = 6.0\,\mathrm{V}$$ $$I_1 = \frac{V_{12}}{R_1}$$ $$I_1 = \frac{6.0}{300}$$ $$I_1 = 0.020\,\mathrm{A}$$

All the current from the top pair flows through the middle wire into the bottom pair.

$$I = I_1 + I_2$$ $$I = 0.020 + 0.010$$ $$I = 0.030\,\mathrm{A}$$

The bottom pair is R3 and R4 in parallel.

$$\frac{1}{R_{34}} = \frac{1}{150} + \frac{1}{300}$$ $$\frac{1}{R_{34}} = 0.0100$$ $$R_{34} = 100\,\Omega$$ $$V_{34} = IR_{34}$$ $$V_{34} = (0.030)(100)$$ $$V_{34} = 3.0\,\mathrm{V}$$

Kirchhoff's voltage law says the battery voltage equals the two drops added together.

$$V = V_{12} + V_{34}$$ $$V = 6.0 + 3.0$$ $$V = 9.0\,\mathrm{V}$$

Example: The bottom wire connects to the battery's positive terminal, the longer line, so let the top wire be at 0 V. What is the electric potential at point A and at point B? What is the potential difference between A and B?

solution

Each column connects straight across the battery, so each column has the full 12 V across it.

$$\text{left column}$$ $$R_{12} = 200 + 400$$ $$R_{12} = 600\,\Omega$$ $$I_{12} = \frac{12}{600}$$ $$I_{12} = 0.020\,\mathrm{A}$$

$$\text{right column}$$ $$R_{34} = 500 + 100$$ $$R_{34} = 600\,\Omega$$ $$I_{34} = \frac{12}{600}$$ $$I_{34} = 0.020\,\mathrm{A}$$

The top wire is at 0 V. Point A is above R1's voltage drop, and point B is above R3's drop.

$$V_A = I_{12}R_1$$ $$V_A = (0.020)(200)$$ $$V_A = 4.0\,\mathrm{V}$$

$$V_B = I_{34}R_3$$ $$V_B = (0.020)(500)$$ $$V_B = 10\,\mathrm{V}$$

$$\Delta V = V_B - V_A$$ $$\Delta V = 10 - 4.0$$ $$\Delta V = 6.0\,\mathrm{V}$$

Both columns carry the same current, but they split the 12 V differently. If you connected a wire from B to A, current would flow from B to A, from high potential to low potential.

Example: A portable speaker uses a 6.0 V battery and draws 0.030 A through its circuit. What power does the battery provide?

solution

This connects Kirchhoff circuit ideas with the power relationship from the previous page.

$$P = IV$$ $$P = (0.030)(6.0)$$ $$P = 0.18\,\mathrm{W}$$

Example: A kitchen outlet circuit runs at 120 V. A toaster (12 Ω) and a microwave (14.4 Ω) are plugged in at the same time, so they are in parallel. The circuit breaker trips above 15 A. Does it trip?

solution

Each appliance gets the full 120 V because they're in parallel.

$$\text{toaster}$$ $$I_1 = \frac{V}{R_1}$$ $$I_1 = \frac{120}{12}$$ $$I_1 = 10\,\mathrm{A}$$

$$\text{microwave}$$ $$I_2 = \frac{V}{R_2}$$ $$I_2 = \frac{120}{14.4}$$ $$I_2 = 8.3\,\mathrm{A}$$

$$I_{\mathrm{in}} = I_{\mathrm{out}}$$ $$I_t = 10 + 8.3$$ $$I_t = 18.3\,\mathrm{A}$$

Yes, 18.3 A is more than 15 A, so the breaker trips. Every appliance you add in parallel adds another path, which lowers the equivalent resistance and increases the total current.

Example: A lighting control board has three parallel indicator branches connected to a 5.0 V supply. The branch currents are 12 mA, 8.0 mA, and 5.0 mA. What total current leaves the supply?

solution

Convert milliamps to amps or add in milliamps first. Here the final answer will be converted to amps.

$$I_t = 12 + 8.0 + 5.0$$ $$I_t = 25$$ $$I_t = 0.025\,\mathrm{A}$$

Example: A 9.0 V battery is connected to a 1.0 kΩ resistor in series with a parallel section. The parallel section has a 2.0 kΩ resistor and a 3.0 kΩ resistor. What is the total current?

solution

$$1.0\,\mathrm{k}\Omega = 1000\,\Omega$$ $$2.0\,\mathrm{k}\Omega = 2000\,\Omega$$ $$3.0\,\mathrm{k}\Omega = 3000\,\Omega$$ $$\frac{1}{R_p} = \frac{1}{2000} + \frac{1}{3000}$$ $$\frac{1}{R_p} = 0.000833$$ $$R_p = 1200$$ $$R_t = 1000 + 1200$$ $$R_t = 2200$$ $$I = \frac{9.0}{2200}$$ $$I = 0.0041\,\mathrm{A}$$

The total current is about 4.1 mA.

Question: In a parallel circuit, why can the current split into different amounts while each branch still has the same voltage across it?

answer

Current can split because charge has more than one path through the junction. The currents in the branches depend on the branch resistances.

The voltage across each branch is the same because each branch connects to the same two points in the circuit. Moving from one side of the branch to the other gives the same potential difference.