1.1 Ohm's Law, Power, and DC Circuit Analysis

Key Takeaways

  • Ohm's law governs DC avionics systems through the fundamental relationships V = I × R, I = V / R, and R = V / I.

  • Electrical power dissipated as heat in resistive circuits is P = V × I = I²R = V² / R; a common design practice derates resistors so they dissipate no more than 50% of their rated wattage.

  • In a series circuit, current is identical everywhere and total resistance equals the sum of individual resistances; in a parallel circuit, voltage is constant across all branches and equivalent resistance is always less than the smallest branch resistance.

  • Kirchhoff's Current Law (KCL) dictates that total current entering a node equals total current exiting (ΣI = 0), while Kirchhoff's Voltage Law (KVL) dictates that the sum of all voltage drops around a closed loop equals source voltage (ΣV = 0).

  • AC 43.13-1B Table 11-6 limits the drop between the bus and the utilization-equipment ground to 1 V continuous (2 V intermittent) on 28 V systems and 0.5 V (1 V) on 14 V systems, about 3.6% of nominal voltage.

Last updated: October 2026

1.1 Ohm's Law, Power, and DC Circuit Analysis

Quick Answer: Direct-current (DC) aircraft electrical systems operate on the governing relationship V=I×RV = I \times R. Electrical power dissipation is determined by P=V×I=I2R=V2RP = V \times I = I^2 R = \frac{V^2}{R}. While series circuits maintain a single constant current path with additive resistances (RT=R1+R2+…R_T = R_1 + R_2 + \dots), aircraft power distribution networks wire loads in parallel so that full bus voltage is supplied across every branch (VT=V1=V2V_T = V_1 = V_2) and the loss of a single component does not interrupt the entire bus. Voltage dividers experience significant output voltage drop ("loading error") whenever an external load resistance RLR_L is connected in parallel with the output resistor. Per FAA AC 43.13-1B Table 11-6, the continuous-operation voltage drop between the bus and the equipment ground should not exceed 1 V on 28V DC systems or 0.5 V on 14V DC systems.


Ohm's Law in Aircraft DC Systems

Every avionics system—from basic cockpit annunciators to integrated primary flight computers—relies on direct-current (DC) power derived from aircraft storage batteries, engine-driven DC generators, or transformer-rectifier units (TRUs). The fundamental mathematical framework describing DC circuit behavior is Ohm's Law, formulated by Georg Simon Ohm. It establishes that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them:

V=I×RV = I \times R

Where:

  • VV is electrical potential difference (Electromotive Force, EMF), measured in Volts (V).
  • II is the intensity of electron flow, measured in Amperes (A), where 1 A=1 Coulomb per second=6.242×1018 electrons/sec1\text{ A} = 1\text{ Coulomb per second} = 6.242 \times 10^{18}\text{ electrons/sec}.
  • RR is the net opposition to electrical current flow, measured in Ohms (Ω\Omega).

By algebraic manipulation, an aircraft electronics technician calculates current or resistance using the two companion forms:

I=VRandR=VII = \frac{V}{R} \quad \text{and} \quad R = \frac{V}{I}

Practical Aircraft Calculation Example

An avionics technician is installing a 28V DC pitot head heater. The technical specifications specify an internal heating element resistance of 5.6 Ω5.6\ \Omega at operating temperature. To calculate the continuous current demand on the DC essential bus:

I=VR=28 V5.6 Ω=5.0 AI = \frac{V}{R} = \frac{28\text{ V}}{5.6\ \Omega} = 5.0\text{ A}

If ground power supplies a reduced bus voltage of 24V during ramp testing, the current drawn decreases proportionally: I=24/5.6=4.29 AI = 24 / 5.6 = 4.29\text{ A}.


Electrical Power and Thermal Derating

Electrical power represents the rate at which electrical energy is converted into work or dissipated as heat. The fundamental power equation states that power (PP) in Watts (W) equals the product of voltage and current:

P=V×IP = V \times I

Substituting Ohm's law (V=I×RV = I \times R and I=V/RI = V / R) yields the three foundational power formulas:

P=V×I=I2R=V2RP = V \times I = I^2 R = \frac{V^2}{R}

Power Derating in Avionics Enclosures

In aircraft electrical installations, heat dissipation is a primary failure mechanism. Resistors dissipate electrical power strictly as thermal energy (I2RI^2 R losses). Operating a resistor at or near its maximum manufacturer-rated wattage causes rapid temperature elevation, resistance value drift, thermal degradation of surrounding printed circuit board (PCB) traces, and premature failure.

A common electronics design practice (not an FAA rule) is to apply a 50% power derating: a resistor should dissipate no more than half of its rated wattage, which matters most in enclosed, poorly ventilated equipment bays. Under that practice, a resistor expected to dissipate 1.0W must be rated for at least 2.0W. When you replace a part, the equipment manufacturer's parts list and derating rules govern.

Minimum Rating≥2×Pcalculated\text{Minimum Rating} \ge 2 \times P_{\text{calculated}}

For example, if a series dropping resistor drops 14V at 0.1A (P=14 V×0.1 A=1.4 WP = 14\text{ V} \times 0.1\text{ A} = 1.4\text{ W}), 50% derating calls for at least 2×1.4=2.8 W2 \times 1.4 = 2.8\text{ W}. A standard 2W part is therefore too small, and the next standard size (3W or larger) is selected.


Series Circuit Analysis

A series circuit provides only a single continuous conducting path for electron flow. As a result, the following rules govern every series circuit:

  1. Current is Constant: The current flowing through every component in the series string is identical: Itotal=I1=I2=I3=⋯=InI_{\text{total}} = I_1 = I_2 = I_3 = \dots = I_n
  2. Resistance is Additive: The total equivalent resistance (RTR_T) equals the arithmetic sum of all individual resistances: Rtotal=R1+R2+R3+⋯+RnR_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n
  3. Voltage Drops are Additive: The total applied voltage (VTV_T) equals the sum of the individual voltage drops across each resistor: Vtotal=V1+V2+V3+⋯+VnV_{\text{total}} = V_1 + V_2 + V_3 + \dots + V_n

If any single component in a series circuit opens, current drops to zero across the entire circuit. For this reason, primary aircraft operational loads are never connected in series.


Parallel Circuit Analysis

A parallel circuit connects two or more components across the same two common electrical nodes, creating multiple independent branch paths for current flow. The governing rules for parallel circuits are:

  1. Voltage is Constant: The voltage drop across every parallel branch is identical and equals the source bus voltage: Vtotal=V1=V2=V3=⋯=VnV_{\text{total}} = V_1 = V_2 = V_3 = \dots = V_n
  2. Branch Currents Add: The total current supplied by the power source equals the sum of the individual branch currents: Itotal=I1+I2+I3+⋯+InI_{\text{total}} = I_1 + I_2 + I_3 + \dots + I_n
  3. Equivalent Resistance Decreases: Total equivalent resistance is always less than the resistance of the smallest individual branch. The reciprocal formula defines parallel resistance: 1Rtotal=1R1+1R2+⋯+1Rn\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}
  4. Conductance (GG): Conductance represents the ease with which current flows, measured in Siemens (S), where G=1/RG = 1 / R. In parallel circuits, total conductance is directly additive: Gtotal=G1+G2+⋯+GnG_{\text{total}} = G_1 + G_2 + \dots + G_n

Two-Resistor Shortcut: Product Over Sum

When exactly two resistors are connected in parallel, technicians utilize the simplified product-over-sum formula:

Rtotal=R1×R2R1+R2R_{\text{total}} = \frac{R_1 \times R_2}{R_1 + R_2}

Numerical Example: An avionics warning panel connects a 60 Ω60\ \Omega lamp in parallel with a 40 Ω40\ \Omega auxiliary relay coil across a 28V DC bus:

Rtotal=60×4060+40=2400100=24 ΩR_{\text{total}} = \frac{60 \times 40}{60 + 40} = \frac{2400}{100} = 24\ \Omega

Total current supplied by the bus:

Itotal=VRtotal=28 V24 Ω=1.167 AI_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{28\text{ V}}{24\ \Omega} = 1.167\text{ A}

Individual branch currents verify Kirchhoff's Current Law:

  • I1=28/60=0.467 AI_1 = 28 / 60 = 0.467\text{ A}
  • I2=28/40=0.700 AI_2 = 28 / 40 = 0.700\text{ A}
  • Itotal=0.467+0.700=1.167 AI_{\text{total}} = 0.467 + 0.700 = 1.167\text{ A}
ParameterSeries CircuitParallel Circuit
Current (II)Constant everywhere: IT=I1=I2I_T = I_1 = I_2Divides across branches: IT=I1+I2I_T = I_1 + I_2
Voltage (VV)Divides across resistors: VT=V1+V2V_T = V_1 + V_2Constant across branches: VT=V1=V2V_T = V_1 = V_2
Total Resistance (RTR_T)Additive: RT=R1+R2R_T = R_1 + R_2 (RT>any branchR_T > \text{any branch})Reciprocal: RT=R1R2R1+R2R_T = \frac{R_1 R_2}{R_1 + R_2} (RT<smallest branchR_T < \text{smallest branch})
Open Fault EffectTotal loss of current to entire stringOnly the affected branch loses power
Aircraft Use CaseInstrument backlighting dimmers, sensor dividersMain avionics buses, flight displays, radios

Kirchhoff's Laws

Complex avionics networks with interconnected loops and distribution buses are analyzed using Gustav Kirchhoff's two governing conservation laws.

Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law (junction rule) is an expression of the conservation of electrical charge. It states that the algebraic sum of currents entering and leaving any circuit node is zero:

∑Ientering=∑Ileaving\sum I_{\text{entering}} = \sum I_{\text{leaving}}

If an aircraft main DC distribution bus node receives 50 A50\text{ A} from Generator 1 and 30 A30\text{ A} from Generator 2, the total current exiting that bus node to various sub-buses and equipment circuit breakers must equal precisely 80 A80\text{ A}.

Kirchhoff's Voltage Law (KVL)

Kirchhoff's Voltage Law (loop rule) is an expression of the conservation of energy. It states that the algebraic sum of all electrical potential differences (source voltages and resistive voltage drops) around any closed loop is zero:

∑Vsources−∑Vdrops=0  ⟹  Vsource=V1+V2+⋯+Vn\sum V_{\text{sources}} - \sum V_{\text{drops}} = 0 \quad \implies \quad V_{\text{source}} = V_1 + V_2 + \dots + V_n

In a 28V DC circuit supplying a navigation receiver through a switch and circuit breaker, if the wire drops 0.4 V0.4\text{ V}, the circuit breaker drops 0.1 V0.1\text{ V}, and the airframe ground return drops 0.2 V0.2\text{ V}, the remaining voltage available directly at the receiver terminals is 28−(0.4+0.1+0.2)=27.3 V28 - (0.4 + 0.1 + 0.2) = 27.3\text{ V}.


Voltage Dividers: Unloaded vs. Loaded

A voltage divider is a series arrangement of two or more resistors used to obtain a specific fractional output voltage from a higher DC supply rail.

Unloaded Voltage Divider

In an open-circuit (unloaded) condition, no current is drawn from the output terminal. The current flowing through both resistors is identical (I=Vin/(R1+R2)I = V_{\text{in}} / (R_1 + R_2)). The output voltage taken across R2R_2 is governed by the voltage divider formula:

Vout=Vin×(R2R1+R2)V_{\text{out}} = V_{\text{in}} \times \left(\frac{R_2}{R_1 + R_2}\right)

Example: An avionics sensor bias network connects R1=18 kΩR_1 = 18\text{ k}\Omega and R2=10 kΩR_2 = 10\text{ k}\Omega in series across a 28V DC bus:

Vout=28 V×(10 kΩ18 kΩ+10 kΩ)=28×(1028)=10.0 VV_{\text{out}} = 28\text{ V} \times \left(\frac{10\text{ k}\Omega}{18\text{ k}\Omega + 10\text{ k}\Omega}\right) = 28 \times \left(\frac{10}{28}\right) = 10.0\text{ V}

The Loading Effect

When an external device with load resistance RLR_L is connected across R2R_2, RLR_L is in parallel with R2R_2. The equivalent lower resistance (ReqR_{\text{eq}}) becomes:

Req=R2×RLR2+RLR_{\text{eq}} = \frac{R_2 \times R_L}{R_2 + R_L}

Because ReqR_{\text{eq}} is strictly less than R2R_2, the loaded output voltage drops below the design voltage:

Vout(loaded)=Vin×(ReqR1+Req)V_{\text{out(loaded)}} = V_{\text{in}} \times \left(\frac{R_{\text{eq}}}{R_1 + R_{\text{eq}}}\right)

Worked Loading Calculation: If a 10 kΩ10\text{ k}\Omega load (RL=10 kΩR_L = 10\text{ k}\Omega) is connected to the divider above:

Req=10 kΩ×10 kΩ10 kΩ+10 kΩ=5.0 kΩR_{\text{eq}} = \frac{10\text{ k}\Omega \times 10\text{ k}\Omega}{10\text{ k}\Omega + 10\text{ k}\Omega} = 5.0\text{ k}\Omega

Vout(loaded)=28 V×(5 kΩ18 kΩ+5 kΩ)=28×(523)=6.09 VV_{\text{out(loaded)}} = 28\text{ V} \times \left(\frac{5\text{ k}\Omega}{18\text{ k}\Omega + 5\text{ k}\Omega}\right) = 28 \times \left(\frac{5}{23}\right) = 6.09\text{ V}

The output voltage sags from 10.0 V10.0\text{ V} down to 6.09 V6.09\text{ V}—a massive 39.1%39.1\% loading error! To minimize loading error in precision avionics circuits, the bleeder current flowing through the divider resistors must be at least 10 times greater than the anticipated load current (Ibleeder≥10×IloadI_{\text{bleeder}} \ge 10 \times I_{\text{load}}), or an active operational-amplifier buffer must isolate the divider.


Circuit Troubleshooting via Voltage Drop Analysis

Voltage drop troubleshooting is the most powerful in-circuit diagnostic method available to an aircraft electronics technician. Rather than disconnecting wires or desoldering components to measure resistance, the technician measures dynamic potential drops while the circuit is energized under normal load.

FAA AC 43.13-1B Allowable Voltage Drop Limits

FAA Advisory Circular AC 43.13-1B, Chapter 11, Section 5 (paragraph 11-66 and Table 11-6) lists the maximum acceptable voltage drop in load circuits between the bus and the utilization equipment ground:

System Nominal VoltageContinuous Operation Max Allowable DropIntermittent Operation Max Allowable Drop
14V DC Bus0.50 V (3.57% of bus)1.00 V (7.14% of bus)
28V DC Bus1.00 V (3.57% of bus)2.00 V (7.14% of bus)
115V AC Bus4.00 V (3.48% of bus)8.00 V (6.96% of bus)

A separate, tighter limit applies upstream of the bus: paragraph 11-66b says the drop in the main power wires from the generator or battery to the bus should not exceed 2 percent of the regulated voltage with the generator carrying rated current. Do not confuse that 2% source-to-bus figure with the Table 11-6 bus-to-load values.

Any conductor, relay contact, or ground bond that pushes a circuit past these limits wastes power as heat and starves the avionics of voltage. AC 43.13-1B adds that if a circuit's measured drop does not exceed the limit set by the aircraft or product manufacturer, its resistance may be considered satisfactory.

Diagnosing Circuit Faults

  1. High-Resistance Fault (Corrosion / Loose Terminals):
    • Symptom: The load operates sluggishly, intermittently, or displays low bus warnings.
    • Method: Connect the digital multimeter (DMM) in DC Volts mode across individual conductors, switches, and ground connections while current is flowing.
    • Interpretation: A good conductor or closed switch displays <0.1 V< 0.1\text{ V}. A corroded terminal block lug, pitted relay contact, or oxidized airframe ground stud will drop significant voltage (e.g., 1.5 V1.5\text{ V} to 5.0 V5.0\text{ V}), starving the downstream load.
  2. Open-Circuit Fault (Broken Wire / Blown Fuse / Open Switch):
    • Symptom: Load is completely inoperative; circuit current is 0.0 A0.0\text{ A}.
    • Interpretation: Across any functional, closed switch or intact fuse, voltage drop is 0.0 V0.0\text{ V}. When measuring directly across an open switch, open contact, or blown fuse in an energized circuit, the meter displays full source voltage (28 V DC28\text{ V DC}). This occurs because the meter's internal 10 MΩ10\text{ M}\Omega input impedance completes the circuit loop to ground, dropping the entire supply voltage.
  3. Short-to-Ground Fault:
    • Symptom: The circuit breaker immediately trips or fuse violently ruptures upon activation.
    • Method: De-energize the circuit, pull the breaker, isolate the load, and measure resistance from the conductor to airframe ground with a DMM in Ohms mode.
    • Interpretation: An ungrounded avionics power lead should measure infinite resistance (>10 MΩ> 10\text{ M}\Omega) to ground. A reading of 0.0 Ω0.0\ \Omega to 0.2 Ω0.2\ \Omega confirms an insulation chafing short against airframe structure.
Test Your Knowledge

In a 28V DC aircraft avionics bus circuit, a relay coil with a resistance of 140 Ω is energized continuously. What is the current draw through the coil, and what is the minimum power rating required for a series dropping resistor if the coil voltage must be dropped to 14V?

A

Current is 2.00 A, and the resistor dissipates 28.0 W (requiring at least a 50 W rated resistor)

B

Current is 0.50 A, and the resistor dissipates 7.00 W (requiring at least a 10 W rated resistor)

C

Current is 0.20 A, and the resistor dissipates 2.80 W (requiring at least a 1 W rated resistor)

D

Current is 0.10 A, and the resistor dissipates 1.40 W (a 3 W part meets 50% derating)

Test Your Knowledge

An avionics technician is measuring an annunciator circuit consisting of two parallel branches connected across a 28V DC bus: Branch 1 has a resistance of 60 Ω and Branch 2 has a resistance of 40 Ω. What is the total equivalent circuit resistance and the total bus current supplied?

A

Total resistance is 12 Ω and total current is 2.33 A

B

Total resistance is 50 Ω and total current is 0.56 A

C

Total resistance is 24 Ω and total current is 1.17 A

D

Total resistance is 100 Ω and total current is 0.28 A

Test Your Knowledge

When troubleshooting a 28V DC landing light circuit where the lamp fails to illuminate with the switch closed, a technician measures 28V DC directly across the closed switch contacts. What does this voltage measurement indicate?

A

The ground return stud has loosened and created a floating ground condition

B

The circuit wiring has a dead short circuit to airframe ground before the switch

C

The switch contacts are open or severely burned, failing to complete the circuit

D

The switch contacts are normal and the filament in the landing light is open

Test Your Knowledge

What occurs when an external resistive load (R_L) is connected across the output resistor (R_2) of an unloaded voltage divider circuit?

A

Output voltage increases because the total circuit resistance decreases, drawing more current

B

Total current drawn from the power source decreases due to current division

C

Output voltage remains unchanged because the voltage divider ratio is fixed by R_1 and R_2

D

Output voltage drops, because R_L in parallel with R_2 lowers the effective resistance of the lower leg

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