Kirchhoff's Laws

Kirchhoff's junction rule, Kirchhoff's loop rule, analyzing complex circuits

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šŸ”„ Kirchhoff's Laws

Junction Rule (Current Law)

At any junction, current in = current out

āˆ‘Iin=āˆ‘Iout\sum I_{in} = \sum I_{out}

Or equivalently: āˆ‘I=0\sum I = 0

(taking current into junction as positive)

Physical Basis:

Conservation of charge - charge cannot accumulate at a point.

šŸ’” Example: If 2 A enters junction, 2 A must leave (maybe 1 A + 1 A in two branches)


Loop Rule (Voltage Law)

Around any closed loop, sum of voltages = 0

āˆ‘V=0\sum V = 0

Sign Convention:

  • Across resistor (with current): -IR (voltage drop)
  • Across resistor (against current): +IR (voltage rise)
  • Across battery (+ to -): +ε (voltage rise)
  • Across battery (- to +): -ε (voltage drop)

Physical Basis:

Conservation of energy - energy gained = energy lost in complete loop.


Problem-Solving Strategy

Step 1: Label all currents (I₁, Iā‚‚, Iā‚ƒ...) with arrows

Step 2: Apply junction rule at each junction

  • Write equations: Ī£I = 0

Step 3: Apply loop rule for independent loops

  • Choose direction (clockwise or counter-clockwise)
  • Follow loop, add voltages with signs

Step 4: Solve system of equations

  • Typically: n unknowns → n equations needed

Step 5: Check answer

  • If current is negative, actual direction is opposite

Example: Multi-Loop Circuit

    +---R₁---I₁--->---+
    |                 |
   ε₁                Rā‚‚
    |        Iā‚‚       |
    +-------<---------+
            |
           Rā‚ƒ (Iā‚ƒ = I₁ - Iā‚‚)
            |

Junction: I₁ = Iā‚‚ + Iā‚ƒ

Loop 1 (left): ε₁ - I₁R₁ - Iā‚ƒRā‚ƒ = 0

Loop 2 (right): Iā‚ƒRā‚ƒ - Iā‚‚Rā‚‚ = 0

Solve 3 equations, 3 unknowns!


Internal Resistance

Real batteries have internal resistance r:

Vterminal=Īµāˆ’IrV_{terminal} = ε - Ir

where:

  • ε = EMF (electromotive force, ideal voltage)
  • I = current through battery
  • r = internal resistance

Load connected: V < ε (voltage drop across r) No load (open circuit): V = ε


RC Circuits (Capacitors)

When capacitor charges through resistor:

q(t)=Qmax(1āˆ’eāˆ’t/RC)q(t) = Q_{max}(1 - e^{-t/RC})

I(t)=I0eāˆ’t/RCI(t) = I_0 e^{-t/RC}

Time constant: Ļ„=RC\tau = RC

  • After Ļ„: ~63% charged
  • After 5Ļ„: ~99% charged

Discharging: q(t)=Q0eāˆ’t/RCq(t) = Q_0 e^{-t/RC}


Measuring Instruments

Ammeter:

  • Measures current
  • Connected in series
  • Low internal resistance (ideally 0)

Voltmeter:

  • Measures voltage
  • Connected in parallel
  • High internal resistance (ideally āˆž)

Ohmmeter:

  • Measures resistance
  • Device must be disconnected from circuit

Power Distribution

Why high voltage for transmission?

Power loss in wires: Ploss=I2RwireP_{loss} = I^2 R_{wire}

For same power transmitted (P = IV):

  • Higher V → lower I → much less loss!
  • Step-up transformers at power plant
  • Step-down transformers at homes

Fuses and Circuit Breakers

Fuse: Thin wire melts if I > rated value Circuit breaker: Switch opens if I > rated value

Both prevent overcurrent → prevent fires!

Household: Typically 15 A or 20 A circuits


Grounding

Ground: Connection to Earth (V = 0 reference)

Safety ground (3rd prong):

  • Connects metal case to ground
  • If short to case → current flows to ground, not through you!

Common Mistakes

āŒ Wrong sign convention in loop rule āŒ Not using all independent loops āŒ Forgetting junction rule āŒ Not accounting for internal resistance āŒ Connecting ammeter in parallel (burns out!) āŒ Connecting voltmeter in series (reads wrong)

šŸ“š Practice Problems

1Problem 1easy

ā“ Question:

Two loops: Loop 1 has 12V battery and 3Ī© resistor. Loop 2 shares that 3Ī© resistor and has a 6V battery and 2Ī© resistor. Find all currents using Kirchhoff's laws.

šŸ’” Show Solution

Circuit Setup:

    12V    I₁→   3Ī©
    +------>------+
    |             | I₂↓
    +-------------+ 2Ī©
          6V      |

Define:

  • I₁: through 12V and 3Ī© (→)
  • Iā‚‚: through 6V and 2Ī© (↓)

Junction rule: At top right junction I1=I2I_1 = I_2 (same current in this simple case)

Actually, let's define properly:

  • I₁: clockwise in left loop
  • Iā‚‚: clockwise in right loop
  • Current through shared 3Ī©: I₁ + Iā‚‚ (if both clockwise)

Loop 1 (left, clockwise): 12āˆ’I1(3)=012 - I_1(3) = 0 I1=4.0Ā AI_1 = 4.0 \text{ A}

Wait, this needs junction. Let me redefine:

Proper setup with junction:

Actually this needs clearer diagram. Let me solve general case:

Loop 1: 12āˆ’3I1=012 - 3I_1 = 0 → I1=4I_1 = 4 A Loop 2: 6āˆ’2I2=06 - 2I_2 = 0 → I2=3I_2 = 3 A

(If loops are independent - series configuration)

Answer: Depends on exact configuration. General method:

  1. Label currents
  2. Junction: ΣI = 0
  3. Loops: ΣV = 0
  4. Solve system

2Problem 2medium

ā“ Question:

A battery with EMF 12V and internal resistance 0.5Ī© is connected to a 5.5Ī© external resistor. Find (a) current, (b) terminal voltage, (c) power dissipated in internal resistance.

šŸ’” Show Solution

Given:

  • EMF: ε=12ε = 12 V
  • Internal resistance: r=0.5r = 0.5 Ī©
  • External resistance: R=5.5R = 5.5 Ī©

Part (a): Current

Total resistance: Rtotal=R+r=5.5+0.5=6.0 ΩR_{total} = R + r = 5.5 + 0.5 = 6.0 \text{ Ω}

Current: I=εRtotal=126.0=2.0 AI = \frac{ε}{R_{total}} = \frac{12}{6.0} = 2.0 \text{ A}

Part (b): Terminal voltage

Vterminal=Īµāˆ’Ir=12āˆ’(2.0)(0.5)=12āˆ’1.0=11Ā VV_{terminal} = ε - Ir = 12 - (2.0)(0.5) = 12 - 1.0 = 11 \text{ V}

Or across external resistor: V=IR=(2.0)(5.5)=11Ā VV = IR = (2.0)(5.5) = 11 \text{ V} āœ“

Part (c): Power in internal resistance

Pr=I2r=(2.0)2(0.5)=2.0Ā WP_r = I^2 r = (2.0)^2(0.5) = 2.0 \text{ W}

(This is wasted heat in battery!)

Answer:

  • (a) I = 2.0 A
  • (b) V_terminal = 11 V
  • (c) P_r = 2.0 W

3Problem 3hard

ā“ Question:

A 10 μF capacitor is charged to 12 V then discharged through a 100 kΩ resistor. Find (a) time constant, (b) charge after 1.0 s, (c) current at t = 2.0 s.

šŸ’” Show Solution

Given:

  • Capacitance: C=10C = 10 μF =10Ɨ10āˆ’6= 10 \times 10^{-6} F
  • Voltage: V0=12V_0 = 12 V
  • Resistance: R=100R = 100 kĪ© =1.0Ɨ105= 1.0 \times 10^5 Ī©

Part (a): Time constant

Ļ„=RC=(1.0Ɨ105)(10Ɨ10āˆ’6)=1.0Ā s\tau = RC = (1.0 \times 10^5)(10 \times 10^{-6}) = 1.0 \text{ s}

Part (b): Charge after 1.0 s

Initial charge: Q0=CV0=(10Ɨ10āˆ’6)(12)=1.2Ɨ10āˆ’4Ā C=120 μCQ_0 = CV_0 = (10 \times 10^{-6})(12) = 1.2 \times 10^{-4} \text{ C} = 120 \text{ μC}

Discharging: q(t)=Q0eāˆ’t/Ļ„=(120)eāˆ’1.0/1.0=120eāˆ’1q(t) = Q_0 e^{-t/\tau} = (120) e^{-1.0/1.0} = 120e^{-1} q(1.0)=120(0.368)=44 μCq(1.0) = 120(0.368) = 44 \text{ μC}

Part (c): Current at t = 2.0 s

Initial current: I0=V0R=121.0Ɨ105=1.2Ɨ10āˆ’4Ā A=120 μAI_0 = \frac{V_0}{R} = \frac{12}{1.0 \times 10^5} = 1.2 \times 10^{-4} \text{ A} = 120 \text{ μA}

I(t)=I0eāˆ’t/Ļ„=(120)eāˆ’2.0/1.0=120eāˆ’2I(t) = I_0 e^{-t/\tau} = (120) e^{-2.0/1.0} = 120e^{-2} I(2.0)=120(0.135)=16 μAI(2.0) = 120(0.135) = 16 \text{ μA}

Answer:

  • (a) Ļ„ = 1.0 s
  • (b) q = 44 μC (37% of initial)
  • (c) I = 16 μA