Kirchhoff's Laws
Kirchhoff's junction rule, Kirchhoff's loop rule, analyzing complex circuits
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š Kirchhoff's Laws
Junction Rule (Current Law)
At any junction, current in = current out
Or equivalently:
(taking current into junction as positive)
Physical Basis:
Conservation of charge - charge cannot accumulate at a point.
š” 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
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 (Iā, Iā, Iā...) with arrows
Step 2: Apply junction rule at each junction
- Write equations: ΣI = 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 ā n equations needed
Step 5: Check answer
- If current is negative, actual direction is opposite
Example: Multi-Loop Circuit
+---Rā---Iā--->---+
| |
εā Rā
| Iā |
+-------<---------+
|
Rā (Iā = Iā - Iā)
|
Junction: Iā = Iā + Iā
Loop 1 (left): εā - IāRā - IāRā = 0
Loop 2 (right): IāRā - IāRā = 0
Solve 3 equations, 3 unknowns!
Internal Resistance
Real batteries have internal resistance r:
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:
Time constant:
- After Ļ: ~63% charged
- After 5Ļ: ~99% charged
Discharging:
Measuring Instruments
Ammeter:
- Measures current
- Connected in series
- Low internal resistance (ideally 0)
Voltmeter:
- Measures voltage
- Connected in parallel
- High internal resistance (ideally ā)
Ohmmeter:
- Measures resistance
- Device must be disconnected from circuit
Power Distribution
Why high voltage for transmission?
Power loss in wires:
For same power transmitted (P = IV):
- Higher V ā lower I ā 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 ā 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 ā current flows to ground, not through you!
Common Mistakes
ā Wrong sign convention in loop rule ā Not using all independent loops ā Forgetting junction rule ā Not accounting for internal resistance ā Connecting ammeter in parallel (burns out!) ā Connecting voltmeter in series (reads wrong)
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.
š” Show Solution
Circuit Setup:
12V Iāā 3Ī©
+------>------+
| | Iāā
+-------------+ 2Ī©
6V |
Define:
- Iā: through 12V and 3Ī© (ā)
- Iā: through 6V and 2Ī© (ā)
Junction rule: At top right junction (same current in this simple case)
Actually, let's define properly:
- Iā: clockwise in left loop
- Iā: clockwise in right loop
- Current through shared 3Ī©: Iā + Iā (if both clockwise)
Loop 1 (left, clockwise):
Wait, this needs junction. Let me redefine:
Proper setup with junction:
Actually this needs clearer diagram. Let me solve general case:
Loop 1: ā A Loop 2: ā A
(If loops are independent - series configuration)
Answer: Depends on exact configuration. General method:
- Label currents
- Junction: ΣI = 0
- Loops: ΣV = 0
- 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.
š” Show Solution
Given:
- EMF: V
- Internal resistance: Ī©
- External resistance: Ī©
Part (a): Current
Total resistance:
Current:
Part (b): Terminal voltage
Or across external resistor: ā
Part (c): Power in internal resistance
(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.
š” Show Solution
Given:
- Capacitance: μF F
- Voltage: V
- Resistance: kΩ Ω
Part (a): Time constant
Part (b): Charge after 1.0 s
Initial charge:
Discharging:
Part (c): Current at t = 2.0 s
Initial current:
Answer:
- (a) Ļ = 1.0 s
- (b) q = 44 μC (37% of initial)
- (c) I = 16 μA