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]
    2A 2A I = ? Example: If a 3-way circuit junction has two wires with 2A each entering. How much current is in the 3rd wire?
    solution $$ I _{in} = I_{out} $$ $$ 2A + 2A = I_{out} $$ $$ 4A = I_{out} $$
    I = ? 20mA 40mA 55mA 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

    $$ I _{in} = I_{out} $$ $$ I _{in} = 20 + 40 + 55 $$ $$ I _{in} = 115 mA $$
    450mA 450mA 115mA I = ? 120mA 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)

    V1 V2 V3 V4

    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]
    1.51V -0.55V -0.33V V = ? Example: What's the voltage drop across the resistor in the diagram?
    solution $$ V_1+V_2+V_3+V_4 = 0 $$ $$ 1.51-0.55-0.33-V_{4} = 0 $$ $$ V_{4} = -0.63V $$
    V = ? 6kΩ 20kΩ 5kΩ 0.194mA 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 V = ? -1.164V -3.88V -0.97V $$ \sum V = 0$$ $$ V_{bat}+V_1+V_2+V_3 = 0 $$ $$ V_{bat} = -V_1-V_2-V_3 $$ $$ V_{bat} = 1.164+3.88+0.97 $$ $$ V_{4} = 6.014V $$

    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.

    R1 R2 R3 R4 Req

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

    \(R_{n}\) = 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]
    100Ω 100Ω 100Ω 100Ω 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.

    R1 R2 R3 Req

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

    \(R_{n}\) = 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 $$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} + \frac{1}{R_{5}}$$ $$\frac{1}{R_{eq}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} + \frac{1}{100} + \frac{1}{100}$$ $$\frac{1}{R_{eq}} = \frac{5}{100}$$ $$R_{eq} = \frac{100}{5}$$ $$R_{eq} = 20 \Omega$$
    400Ω 200Ω 100Ω Example: Find the equivalent resistance for the three resistors above.
    solution $$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$$ $$\frac{1}{R_{eq}} = \frac{1}{400} + \frac{1}{200} + \frac{1}{100}$$ $$\frac{1}{R_{eq}} = \frac{1}{400} + \left(\frac{2}{2}\right)\frac{1}{200} + \left(\frac{4}{4}\right)\frac{1}{100}$$ $$\frac{1}{R_{eq}} = \frac{1}{400} + \frac{2}{400} + \frac{4}{400}$$ $$\frac{1}{R_{eq}} = \frac{7}{400}$$ $$R_{eq} = \frac{400}{7}$$ $$R_{eq} = 57.14 \Omega$$
    30kΩ R2 = ? 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.

    6.0V R1 600Ω R2 800Ω R3 900Ω R4 500Ω

    We can start by combining the two parallel resistors.

    $$\frac{1}{R_{eq}} = \frac{1}{R_{2}} + \frac{1}{R_{3}}$$ $$R_{eq} = \left(\frac{1}{800} + \frac{1}{900}\right)^{-1}$$ $$R_{eq} = \left(0.00125 + 0.00\overline{11}\right)^{-1}$$ $$R_{eq} = 420 \Omega$$ 6.0V R1 600Ω Req 420Ω R4 500Ω

    Next we can combine the 3 resistors in series.

    $$R_{eq} = 600+420+500$$ $$R_{eq} = 1520 \Omega$$ 6.0V Req 1520Ω 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.

    $$V_{bat} + V_1 = 0$$ $$6.0 + V_1 = 0$$ $$V_1 = -6.0$$ 6.0V Req 1520Ω -6.0V

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

    $$V=IR$$ $$I = \frac{V}{R}$$ $$I = \frac{6.0}{1530}$$ $$I = 0.0039A$$ 6.0V Req 0.0039A 1523Ω -6.0V

    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.

    6.0V R1 600Ω 0.0039A Req 420Ω 0.0039A R3 900Ω R4 500Ω 0.0039A

    We can use Ohm's law to find voltages.

    $$V=IR_1$$ $$V=(0.0039)(600)$$ $$V=2.3V$$
    $$V=IR_4$$ $$V=(0.0039)(500)$$ $$V=2.0V$$
    $$V=IR_{eq}$$ $$V=(0.0039)(420)$$ $$V=1.7V$$ 6.0V R1 600Ω 0.0039A -2.3V Req 420Ω 0.0039A -1.7V R3 900Ω R4 500Ω 0.0039A -2.0V

    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.

    6.0V R1 600Ω 0.0039A -2.3V R2 800Ω -1.7V R3 900Ω -1.7V R4 500Ω 0.0039A -2.0V

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

    $$I=\frac{V}{R_2}$$ $$I=\frac{1.7}{800}$$ $$I=0.0021A$$
    $$I=\frac{V}{R_3}$$ $$I=\frac{1.7}{900}$$ $$I=0.0019A$$
    6.0V R1 600Ω 0.0039A -2.3V R2 800Ω -1.7V 0.0021A R3 900Ω -1.7V 0.0019A R4 500Ω 0.0039A -2.0V

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

    $$P = IV_1$$ $$P = (0.0039)(2.3)$$ $$P = 0.0090W$$
    $$P = IV_4$$ $$P = (0.0039)(2.0)$$ $$P = 0.0078W$$
    $$P = IV_2$$ $$P = (0.0021)(1.7)$$ $$P = 0.0036W$$
    $$P = IV_3$$ $$P = (0.0019)(1.7)$$ $$P = 0.0032W$$
    6.0V R1 600Ω 0.0039A -2.3V 0.0090W R2 800Ω -1.7V 0.0021A 0.0036W R3 900Ω -1.7V 0.0019A 0.0032W R4 500Ω 0.0039A -2.0V 0.0078W

    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!

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