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7.2 Circuit Analysis: Series, Parallel, and Combination Circuits with Kirchhoff's Laws
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7.2 Circuit Analysis: Series, Parallel, and Combination Circuits with Kirchhoff's Laws
Key Takeaways Explain This Topic AI
In a series circuit, current is identical through all components ( I T = I 1 = I 2 I_T = I_1 = I_2 I T = I 1 = I 2 ), total resistance is strictly additive ( R T = ∑ R n R_T = \sum R_n R T = ∑ R n ), and total voltage equals the sum of individual voltage drops.
In a parallel circuit, voltage is identical across all branches ( V T = V 1 = V 2 V_T = V_1 = V_2 V T = V 1 = V 2 ), total current equals the sum of branch currents ( I T = ∑ I n I_T = \sum I_n I T = ∑ I n ), and equivalent resistance follows the reciprocal formula 1 R T = ∑ 1 R n \frac{1}{R_T} = \sum \frac{1}{R_n} R T 1 = ∑ R n 1 .
For two parallel resistors, the shortcut is R T = R 1 R 2 R 1 + R 2 R_T = \frac{R_1 R_2}{R_1 + R_2} R T = R 1 + R 2 R 1 R 2 ; for n n n identical parallel resistors, R T = R n R_T = \frac{R}{n} R T = n R ; equivalent parallel resistance is always strictly less than the smallest branch resistance.
Combination (series-parallel) networks are solved by systematically reducing pure parallel and series sub-clusters into single equivalent resistances until one total circuit resistance remains.
Kirchhoff's Current Law (KCL, ∑ I i n = ∑ I o u t \sum I_{in} = \sum I_{out} ∑ I in = ∑ I o u t ) enforces conservation of charge at nodes, while Kirchhoff's Voltage Law (KVL, ∑ V s o u r c e s − ∑ V d r o p s = 0 \sum V_{sources} - \sum V_{drops} = 0 ∑ V so u rces − ∑ V d ro p s = 0 ) enforces conservation of energy around closed loops.
Last updated: August 2026
7.2 Circuit Analysis: Series, Parallel, and Combination Circuits with Kirchhoff's Laws
Core Principle: Electrical and electronic systems route power and signals through three primary network configurations: series , parallel , and combination (series-parallel) topologies. Rigorous circuit analysis relies on combining Ohm's Law with Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) to calculate total equivalent resistance, determine branch currents, analyze voltage drops, and isolate circuit faults.
Fundamental Circuit States
Before analyzing multi-component networks, every military technician must evaluate the three operational states of an electrical circuit: +-----------------------------------------------------------------------------------------+
- | THREE CORE CIRCUIT STATES |
- +-------------------+-----------------+-------------------+-------------------------------+
- | Circuit State | Resistance (R) | Current Flow (I) | Voltage Characteristics |
- +-------------------+-----------------+-------------------+-------------------------------+
- | 1. Closed Circuit | Normal load R | Normal (I = V / R)| Normal voltage drops across |
- | (Complete) | (e.g., 10-100 Ω)| | individual resistive loads |
- +-------------------+-----------------+-------------------+-------------------------------+
- | 2. Open Circuit | Infinite | Zero | Full power source voltage |
- | (Broken Path) | (R = ∞) | (I = 0) | appears across the open gap |
- +-------------------+-----------------+-------------------+-------------------------------+
- | 3. Short Circuit | Near Zero | Destructive surge | Voltage across shorted load |
- | (Bypass Path) | (R ≈ 0 Ω) | (I → ∞) | drops to 0V; trips breakers |
- +-------------------+-----------------+-------------------+-------------------------------+
Closed (Complete) Circuit: An unbroken conductive pathway exists from the positive terminal of the power supply, through the control switches and resistive loads, and back to the negative return terminal. Current flows smoothly in accordance with Ohm's Law.
Open Circuit: An unintentional physical discontinuity, broken wire, opened switch, or blown fuse interrupts the conductive path. Because the air gap presents infinite resistance ( R = ∞ R = \infty R = ∞ ) , current flow immediately drops to zero ( I = 0 A I = 0\text{ A} I = 0 A ) . Crucially, the full source EMF appears directly across the open terminals , which can present a severe electrical shock hazard to maintenance personnel testing the circuit.
Short Circuit: An accidental zero-resistance conductive bridge bypasses the intended load resistance. Because total circuit resistance collapses toward zero ( R ≈ 0 Ω R \approx 0\,\Omega R ≈ 0 Ω ), current surges exponentially ( I = V / R → ∞ I = V / R \rightarrow \infty I = V / R → ∞ ). This extreme overcurrent generates intense I 2 R I^2 R I 2 R thermal energy within milliseconds, vaporizing wiring insulation, melting connectors, and starting electrical fires unless cleared instantly by a fuse or circuit breaker.
Series Circuit Architecture & Governing Laws
A series circuit provides exactly one continuous path for electric current. All circuit elements are connected end-to-end in a single string. +-------------------[ R1 ]---[ R2 ]---[ R3 ]-------------------+
- | |
- [ + ] |
- (Source) |
- [ - ] |
+--------------------------------------------------------------+ +-----------------------------------------------------------------------------------------+
- | SERIES CIRCUIT GOVERNING LAWS |
- +-------------------+-----------------------------+---------------------------------------+
- | Circuit Parameter | Mathematical Formula | Physical Consequence & Principle |
- +-------------------+-----------------------------+---------------------------------------+
- | Current (I) | I_T = I₁ = I₂ = I₃ = ... | Current is identical at all points; |
- | | | charge cannot accumulate or escape. |
- +-------------------+-----------------------------+---------------------------------------+
- | Resistance (R) | R_T = R₁ + R₂ + R₃ + ... | Strictly additive; total R is always |
- | | | GREATER than largest single resistor. |
- +-------------------+-----------------------------+---------------------------------------+
- | Voltage (V) | V_T = V₁ + V₂ + V₃ + ... | Total source EMF equals the algebraic |
- | | | sum of all individual voltage drops. |
- +-------------------+-----------------------------+---------------------------------------+
- | Power (P) | P_T = P₁ + P₂ + P₃ + ... | Total power is the direct sum of all |
- | | | individual component dissipations. |
- +-------------------+-----------------------------+---------------------------------------+
- The Voltage Divider Rule
Because the identical current I T I_T I T passes through each series resistor, the voltage drop across any individual resistor R x R_x R x is directly proportional to its resistance value relative to the total circuit resistance R T R_T R T : V x = V T ⋅ ( R x R T ) V_x = V_T \cdot \left(\frac{R_x}{R_T}\right) V x = V T ⋅ ( R T R x )
Comprehensive Series Circuit Worked Example
Given Parameters: A 48.0 V 48.0\text{ V} 48.0 V DC tactical vehicle power supply is connected across three series resistors: R 1 = 6.0 Ω R_1 = 6.0\,\Omega R 1 = 6.0 Ω , R 2 = 10.0 Ω R_2 = 10.0\,\Omega R 2 = 10.0 Ω , and R 3 = 8.0 Ω R_3 = 8.0\,\Omega R 3 = 8.0 Ω .
Step 1: Calculate Total Series Resistance ( R T R_T R T ): R T = R 1 + R 2 + R 3 = 6.0 Ω + 10.0 Ω + 8.0 Ω = 24.0 Ω R_T = R_1 + R_2 + R_3 = 6.0\,\Omega + 10.0\,\Omega + 8.0\,\Omega = 24.0\,\Omega R T = R 1 + R 2 + R 3 = 6.0 Ω + 10.0 Ω + 8.0 Ω = 24.0 Ω
Step 2: Calculate Circuit Current ( I T I_T I T ): I T = V T R T = 48.0 V 24.0 Ω = 2.0 A I_T = \frac{V_T}{R_T} = \frac{48.0\text{ V}}{24.0\,\Omega} = 2.0\text{ A} I T = R T V T = 24.0 Ω 48.0 V = 2.0 A (Note: Exactly 2.0 A 2.0\text{ A} 2.0 A flows through R 1 R_1 R 1 , R 2 R_2 R 2 , and R 3 R_3 R 3 .)
Step 3: Calculate Individual Voltage Drops ( V n = I T ⋅ R n V_n = I_T \cdot R_n V n = I T ⋅ R n ): V 1 = I T ⋅ R 1 = 2.0 A × 6.0 Ω = 12.0 V V_1 = I_T \cdot R_1 = 2.0\text{ A} \times 6.0\,\Omega = 12.0\text{ V} V 1 = I T ⋅ R 1 = 2.0 A × 6.0 Ω = 12.0 V V 2 = I T ⋅ R 2 = 2.0 A × 10.0 Ω = 20.0 V V_2 = I_T \cdot R_2 = 2.0\text{ A} \times 10.0\,\Omega = 20.0\text{ V} V 2 = I T ⋅ R 2 = 2.0 A × 10.0 Ω = 20.0 V V 3 = I T ⋅ R 3 = 2.0 A × 8.0 Ω = 16.0 V V_3 = I_T \cdot R_3 = 2.0\text{ A} \times 8.0\,\Omega = 16.0\text{ V} V 3 = I T ⋅ R 3 = 2.0 A × 8.0 Ω = 16.0 V
Step 4: Verify via Kirchhoff's Voltage Law (KVL): V T = V 1 + V 2 + V 3 = 12.0 V + 20.0 V + 16.0 V = 48.0 V (Matches source) V_T = V_1 + V_2 + V_3 = 12.0\text{ V} + 20.0\text{ V} + 16.0\text{ V} = 48.0\text{ V} \quad \text{(Matches source)} V T = V 1 + V 2 + V 3 = 12.0 V + 20.0 V + 16.0 V = 48.0 V (Matches source)
Step 5: Calculate Total Power Dissipation ( P T P_T P T ): P T = V T ⋅ I T = 48.0 V × 2.0 A = 96.0 W P_T = V_T \cdot I_T = 48.0\text{ V} \times 2.0\text{ A} = 96.0\text{ W} P T = V T ⋅ I T = 48.0 V × 2.0 A = 96.0 W P 1 = I T 2 R 1 = ( 2.0 ) 2 × 6.0 = 24.0 W ; P 2 = ( 2.0 ) 2 × 10.0 = 40.0 W ; P 3 = ( 2.0 ) 2 × 8.0 = 32.0 W P_1 = I_T^2 R_1 = (2.0)^2 \times 6.0 = 24.0\text{ W}; \quad P_2 = (2.0)^2 \times 10.0 = 40.0\text{ W}; \quad P_3 = (2.0)^2 \times 8.0 = 32.0\text{ W} P 1 = I T 2 R 1 = ( 2.0 ) 2 × 6.0 = 24.0 W ; P 2 = ( 2.0 ) 2 × 10.0 = 40.0 W ; P 3 = ( 2.0 ) 2 × 8.0 = 32.0 W P T = 24.0 W + 40.0 W + 32.0 W = 96.0 W P_T = 24.0\text{ W} + 40.0\text{ W} + 32.0\text{ W} = 96.0\text{ W} P T = 24.0 W + 40.0 W + 32.0 W = 96.0 W
Series Failure Mode Rule: If any single component in a series string fails in an open state (such as a burned filament in older decorative holiday lights), total circuit current drops to zero, and all components cease operating immediately.
Parallel Circuit Architecture & Governing Laws
A parallel circuit connects two or more electrical branches across the same two common electrical nodes, providing multiple independent pathways for current flow. +---------+---------+---------+
- | | | |
- | [ R1 ] [ R2 ] [ R3 ]
- [ + ] | | |
- (Source) | | |
- [ - ] | | |
- | | | |
+---------+---------+---------+ +-----------------------------------------------------------------------------------------+
- | PARALLEL CIRCUIT GOVERNING LAWS |
- +-------------------+-----------------------------+---------------------------------------+
- | Circuit Parameter | Mathematical Formula | Physical Consequence & Principle |
- +-------------------+-----------------------------+---------------------------------------+
- | Voltage (V) | V_T = V₁ = V₂ = V₃ = ... | Every parallel branch connects |
- | | | directly across the full source EMF. |
- +-------------------+-----------------------------+---------------------------------------+
- | Current (I) | I_T = I₁ + I₂ + I₃ + ... | Total current drawn equals the sum of |
- | | | all individual branch currents (KCL). |
- +-------------------+-----------------------------+---------------------------------------+
- | Resistance (R) | 1/R_T = 1/R₁ + 1/R₂ + ... | Reciprocal sum; total equivalent R is |
- | | | always LESS than the smallest branch. |
- +-------------------+-----------------------------+---------------------------------------+
- | Power (P) | P_T = P₁ + P₂ + P₃ + ... | Total power is the direct sum of all |
- | | | branch power dissipations. |
- +-------------------+-----------------------------+---------------------------------------+
- Parallel Resistance Formulas & Calculation Shortcuts
General Reciprocal Formula (Applicable to any number of parallel branches): 1 R T = 1 R 1 + 1 R 2 + 1 R 3 + ⋯ + 1 R n ⟹ R T = 1 ∑ k = 1 n 1 R k \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n} \implies R_T = \frac{1}{\sum_{k=1}^n \frac{1}{R_k}} R T 1 = R 1 1 + R 2 1 + R 3 1 + ⋯ + R n 1 ⟹ R T = ∑ k = 1 n R k 1 1
Product-Over-Sum Shortcut (Exclusively for EXACTLY TWO parallel resistors): R T = R 1 ⋅ R 2 R 1 + R 2 R_T = \frac{R_1 \cdot R_2}{R_1 + R_2} R T = R 1 + R 2 R 1 ⋅ R 2
Calculation Example: Two parallel resistors of 20.0 Ω 20.0\,\Omega 20.0 Ω and 30.0 Ω 30.0\,\Omega 30.0 Ω : R T = 20.0 × 30.0 20.0 + 30.0 = 600.0 50.0 = 12.0 Ω R_T = \frac{20.0 \times 30.0}{20.0 + 30.0} = \frac{600.0}{50.0} = 12.0\,\Omega R T = 20.0 + 30.0 20.0 × 30.0 = 50.0 600.0 = 12.0 Ω
Identical Resistors in Parallel Shortcut: When n n n parallel resistors possess the identical resistance value R R R : R T = R n R_T = \frac{R}{n} R T = n R
Calculation Example: Five 100.0 Ω 100.0\,\Omega 100.0 Ω resistors in parallel yield: R T = 100.0 Ω 5 = 20.0 Ω R_T = \frac{100.0\,\Omega}{5} = 20.0\,\Omega R T = 5 100.0 Ω = 20.0 Ω
The Golden Parallel Resistance Law: The equivalent resistance ( R T R_T R T ) of a parallel network is ALWAYS strictly less than the resistance of the smallest individual branch resistor . If a 10 Ω 10\,\Omega 10 Ω resistor is placed in parallel with a 100,000 Ω 100{,}000\,\Omega 100 , 000 Ω resistor, the total resistance must be less than 10 Ω 10\,\Omega 10 Ω ( R T ≈ 9.999 Ω R_T \approx 9.999\,\Omega R T ≈ 9.999 Ω ).
Current Divider Rule (Two Parallel Branches)
Because each branch experiences the same voltage, current divides inversely proportional to branch resistance—the branch with lower resistance draws a proportionally larger share of total current: I 1 = I T ⋅ ( R 2 R 1 + R 2 ) and I 2 = I T ⋅ ( R 1 R 1 + R 2 ) I_1 = I_T \cdot \left(\frac{R_2}{R_1 + R_2}\right) \quad \text{and} \quad I_2 = I_T \cdot \left(\frac{R_1}{R_1 + R_2}\right) I 1 = I T ⋅ ( R 1 + R 2 R 2 ) and I 2 = I T ⋅ ( R 1 + R 2 R 1 )
Comprehensive Parallel Circuit Worked Example
Given Parameters: A 36.0 V 36.0\text{ V} 36.0 V DC power supply connected across three parallel branches: R 1 = 18.0 Ω R_1 = 18.0\,\Omega R 1 = 18.0 Ω , R 2 = 9.0 Ω R_2 = 9.0\,\Omega R 2 = 9.0 Ω , and R 3 = 6.0 Ω R_3 = 6.0\,\Omega R 3 = 6.0 Ω .
Step 1: Calculate Equivalent Parallel Resistance ( R T R_T R T ): 1 R T = 1 18.0 + 1 9.0 + 1 6.0 = 1 18.0 + 2 18.0 + 3 18.0 = 6.0 18.0 = 1 3.0 ⟹ R T = 3.0 Ω \frac{1}{R_T} = \frac{1}{18.0} + \frac{1}{9.0} + \frac{1}{6.0} = \frac{1}{18.0} + \frac{2}{18.0} + \frac{3}{18.0} = \frac{6.0}{18.0} = \frac{1}{3.0} \implies R_T = 3.0\,\Omega R T 1 = 18.0 1 + 9.0 1 + 6.0 1 = 18.0 1 + 18.0 2 + 18.0 3 = 18.0 6.0 = 3.0 1 ⟹ R T = 3.0 Ω
Step 2: Calculate Individual Branch Currents ( I n = V T / R n I_n = V_T / R_n I n = V T / R n ): I 1 = 36.0 V 18.0 Ω = 2.0 A I_1 = \frac{36.0\text{ V}}{18.0\,\Omega} = 2.0\text{ A} I 1 = 18.0 Ω 36.0 V = 2.0 A I 2 = 36.0 V 9.0 Ω = 4.0 A I_2 = \frac{36.0\text{ V}}{9.0\,\Omega} = 4.0\text{ A} I 2 = 9.0 Ω 36.0 V = 4.0 A I 3 = 36.0 V 6.0 Ω = 6.0 A I_3 = \frac{36.0\text{ V}}{6.0\,\Omega} = 6.0\text{ A} I 3 = 6.0 Ω 36.0 V = 6.0 A
Step 3: Calculate Total Circuit Current ( I T I_T I T ): I T = V T R T = 36.0 V 3.0 Ω = 12.0 A I_T = \frac{V_T}{R_T} = \frac{36.0\text{ V}}{3.0\,\Omega} = 12.0\text{ A} I T = R T V T = 3.0 Ω 36.0 V = 12.0 A Verification (KCL): I T = I 1 + I 2 + I 3 = 2.0 A + 4.0 A + 6.0 A = 12.0 A I_T = I_1 + I_2 + I_3 = 2.0\text{ A} + 4.0\text{ A} + 6.0\text{ A} = 12.0\text{ A} I T = I 1 + I 2 + I 3 = 2.0 A + 4.0 A + 6.0 A = 12.0 A .
Parallel Failure Mode Rule: If one branch in a parallel circuit opens, the remaining branches continue operating completely normally because their branch voltage remains unchanged. This is why civilian household wiring and military vehicle subsystems are wired in parallel.
Combination (Series-Parallel) Circuits: Network Reduction
Most operational electronic systems combine series and parallel configurations on a single circuit board. Analyzing combination networks requires the Equivalent Resistance Network Reduction Method . +-----------------------------------------------------------------------------------------+
- | COMBINATION CIRCUIT REDUCTION ALGORITHM |
- +-----------------------------------------------------------------------------------------+
- | Step 1: Scan the schematic and identify pure parallel or pure series sub-clusters. |
- | Step 2: Replace each sub-cluster with its calculated single equivalent resistance. |
- | Step 3: Redraw the simplified schematic diagram. |
- | Step 4: Repeat the reduction loop until a single total equivalent resistance RT remains.|
- | Step 5: Solve for total source current: IT = VT / RT. |
- | Step 6: Work backward through each node to calculate branch voltages and currents. |
+-----------------------------------------------------------------------------------------+ ORIGINAL COMBINATION NETWORK:
- +---[ R1: 15Ω ]---+-------+---[ R2: 20Ω ]---+-------+
- | | | | |
- [ + ] | +---[ R3: 30Ω ]---+ |
- ( 72V ) | |
- [ - ] +---------------------------------+
- | |
- +---------------------------------------------------+
STEP 1: REDUCE PARALLEL CLUSTER (R2 || R3):
R23 = (20 × 30) / (20 + 30) = 600 / 50 = 12Ω
STEP 2: EQUIVALENT SERIES REDUCTION:
- +---[ R1: 15Ω ]---[ R23: 12Ω ]---+
- | |
- (72V) |
- | |
- +--------------------------------+
- RT = R1 + R23 = 15Ω + 12Ω = 27Ω
- Comprehensive Worked Combination Example
Given Parameters: A 72.0 V 72.0\text{ V} 72.0 V DC supply powering a combination circuit: Resistor R 1 = 15.0 Ω R_1 = 15.0\,\Omega R 1 = 15.0 Ω is in series with a parallel bank consisting of R 2 = 20.0 Ω R_2 = 20.0\,\Omega R 2 = 20.0 Ω and R 3 = 30.0 Ω R_3 = 30.0\,\Omega R 3 = 30.0 Ω .
Step 1: Simplify Parallel Bank ( R 23 R_{23} R 23 ): R 23 = R 2 ⋅ R 3 R 2 + R 3 = 20.0 × 30.0 20.0 + 30.0 = 600.0 50.0 = 12.0 Ω R_{23} = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{20.0 \times 30.0}{20.0 + 30.0} = \frac{600.0}{50.0} = 12.0\,\Omega R 23 = R 2 + R 3 R 2 ⋅ R 3 = 20.0 + 30.0 20.0 × 30.0 = 50.0 600.0 = 12.0 Ω
Step 2: Calculate Total Circuit Resistance ( R T R_T R T ): R T = R 1 + R 23 = 15.0 Ω + 12.0 Ω = 27.0 Ω R_T = R_1 + R_{23} = 15.0\,\Omega + 12.0\,\Omega = 27.0\,\Omega R T = R 1 + R 23 = 15.0 Ω + 12.0 Ω = 27.0 Ω
Step 3: Calculate Total Source Current ( I T I_T I T ): I T = V S R T = 72.0 V 27.0 Ω = 2.667 A = 8 3 A I_T = \frac{V_S}{R_T} = \frac{72.0\text{ V}}{27.0\,\Omega} = 2.667\text{ A} = \frac{8}{3}\text{ A} I T = R T V S = 27.0 Ω 72.0 V = 2.667 A = 3 8 A
Step 4: Calculate Voltage Drop Across Series Resistor R 1 R_1 R 1 : V 1 = I T ⋅ R 1 = 2.667 A × 15.0 Ω = 40.0 V V_1 = I_T \cdot R_1 = 2.667\text{ A} \times 15.0\,\Omega = 40.0\text{ V} V 1 = I T ⋅ R 1 = 2.667 A × 15.0 Ω = 40.0 V
Step 5: Calculate Voltage Across Parallel Bank ( V p a r a l l e l V_{parallel} V p a r a ll e l ): By KVL: V p a r a l l e l = V S − V 1 = 72.0 V − 40.0 V = 32.0 V V_{parallel} = V_S - V_1 = 72.0\text{ V} - 40.0\text{ V} = 32.0\text{ V} V p a r a ll e l = V S − V 1 = 72.0 V − 40.0 V = 32.0 V (Or: V p a r a l l e l = I T ⋅ R 23 = 2.667 A × 12.0 Ω = 32.0 V V_{parallel} = I_T \cdot R_{23} = 2.667\text{ A} \times 12.0\,\Omega = 32.0\text{ V} V p a r a ll e l = I T ⋅ R 23 = 2.667 A × 12.0 Ω = 32.0 V )
Step 6: Calculate Individual Branch Currents: I 2 = V p a r a l l e l R 2 = 32.0 V 20.0 Ω = 1.60 A I_2 = \frac{V_{parallel}}{R_2} = \frac{32.0\text{ V}}{20.0\,\Omega} = 1.60\text{ A} I 2 = R 2 V p a r a ll e l = 20.0 Ω 32.0 V = 1.60 A I 3 = V p a r a l l e l R 3 = 32.0 V 30.0 Ω = 1.067 A I_3 = \frac{V_{parallel}}{R_3} = \frac{32.0\text{ V}}{30.0\,\Omega} = 1.067\text{ A} I 3 = R 3 V p a r a ll e l = 30.0 Ω 32.0 V = 1.067 A Verification (KCL): I 2 + I 3 = 1.60 A + 1.067 A = 2.667 A = I T I_2 + I_3 = 1.60\text{ A} + 1.067\text{ A} = 2.667\text{ A} = I_T I 2 + I 3 = 1.60 A + 1.067 A = 2.667 A = I T .
Kirchhoff's Laws of Circuit Analysis
Formulated by Gustav Kirchhoff in 1845, these two fundamental physical laws form the universal foundation for all electrical engineering network calculations: +-----------------------------------------------------------------------------------------+
- | KIRCHHOFF'S TWO LAWS |
- +-------------------+--------------------------+------------------------------------------+
- | Law Designation | Fundamental Equation | Governing Physical Conservation Law |
- +-------------------+--------------------------+------------------------------------------+
- | 1. Current Law | Σ I_in = Σ I_out | Conservation of Electric Charge: Charge |
- | (KCL / Node) | or Σ I_node = 0 | cannot be created or destroyed at a node.|
- +-------------------+--------------------------+------------------------------------------+
- | 2. Voltage Law | Σ V_sources = Σ V_drops | Conservation of Energy: Total electrical |
- | (KVL / Loop) | or Σ ΔV_loop = 0 | energy gained equals energy dissipated. |
- +-------------------+--------------------------+------------------------------------------+
- 1. Kirchhoff's Current Law (KCL — The Node / Junction Rule)
At any electrical junction or node where two or more conductors meet, electric charge cannot accumulate, store, or vanish. Therefore, the total electric current entering a node must precisely equal the total electric current exiting the node : ∑ I i n = ∑ I o u t ⟺ ∑ I n o d e = 0 \sum I_{in} = \sum I_{out} \quad \Longleftrightarrow \quad \sum I_{node} = 0 ∑ I in = ∑ I o u t ⟺ ∑ I n o d e = 0
Node Problem Example: A circuit node connects four wires:
Wire A carries 5.5 A 5.5\text{ A} 5.5 A into the node.
Wire B carries 3.0 A 3.0\text{ A} 3.0 A into the node.
Wire C carries 4.2 A 4.2\text{ A} 4.2 A away from the node.
What is the current in Wire D?
Solution: ∑ I i n = ∑ I o u t ⟹ 5.5 A + 3.0 A = 4.2 A + I D ⟹ 8.5 A = 4.2 A + I D ⟹ I D = 4.3 A \sum I_{in} = \sum I_{out} \implies 5.5\text{ A} + 3.0\text{ A} = 4.2\text{ A} + I_D \implies 8.5\text{ A} = 4.2\text{ A} + I_D \implies I_D = 4.3\text{ A} ∑ I in = ∑ I o u t ⟹ 5.5 A + 3.0 A = 4.2 A + I D ⟹ 8.5 A = 4.2 A + I D ⟹ I D = 4.3 A exiting the node.
2. Kirchhoff's Voltage Law (KVL — The Closed Loop Rule)
Around any complete, closed conductive loop in an electrical network, energy conservation dictates that the algebraic sum of all electromotive force gains (power sources) minus all potential difference drops (resistive loads) must equal zero : ∑ V s o u r c e s − ∑ V d r o p s = 0 ⟺ ∑ Δ V l o o p = 0 \sum V_{sources} - \sum V_{drops} = 0 \quad \Longleftrightarrow \quad \sum \Delta V_{loop} = 0 ∑ V so u rces − ∑ V d ro p s = 0 ⟺ ∑ Δ V l oo p = 0
Loop Analysis Sign Conventions:
Tracing through a DC voltage source from negative ( − - − ) to positive ( + + + ): + V +V + V (potential gain).
Tracing through a DC voltage source from positive ( + + + ) to negative ( − - − ) : − V -V − V (potential drop).
Tracing across a resistor in the direction of current flow: − I ⋅ R -I \cdot R − I ⋅ R (potential drop).
Tracing across a resistor against the direction of current flow: + I ⋅ R +I \cdot R + I ⋅ R (potential gain).