AcademyDC Circuits
Academy
Power Distribution
Level 1 - Physics topic page in DC Circuits.
Principle
Power distribution is the accounting of energy transfer in a circuit. The same delivered power can require very different currents, and current determines resistive loss.
Notation
\(P\)
electrical power
\(\mathrm{W}\)
\(I\)
current
\(\mathrm{A}\)
\(V\)
potential difference
\(\mathrm{V}\)
\(R_{\mathrm{line}}\)
total resistance of supply and return conductors
\(\mathrm{\Omega}\)
\(R_L\)
load resistance
\(\mathrm{\Omega}\)
\(\eta\)
efficiency
unitless
Method
Derivation 1: Power in a circuit element
Power is energy transferred per unit time. A current moves charge through a potential difference, so the power associated with an element is current times voltage.
Energy per charge
\[\Delta U=qV\]
Power
\[P=\frac{dU}{dt}=V\frac{dq}{dt}\]
Circuit form
\[P=IV\]
Resistor forms
\[P=I^2R=\frac{V^2}{R}\]
Derivation 2: Line losses
Transmission wires have resistance. If the same current passes through the line resistance, the line converts electrical energy into thermal energy.
Line voltage drop
\[\Delta V_{\mathrm{line}}=IR_{\mathrm{line}}\]
Line loss
\[P_{\mathrm{loss}}=I^2R_{\mathrm{line}}\]
Current for delivered power
\[I\approx\frac{P_{\mathrm{delivered}}}{V_{\mathrm{delivery}}}\]
Loss scaling
\[P_{\mathrm{loss}}\approx\frac{P_{\mathrm{delivered}}^2R_{\mathrm{line}}}{V_{\mathrm{delivery}}^2}\]
Derivation 3: Source resistance and load power
A real source can be modeled as an ideal emf in series with internal resistance. The load power is not the same as the source power because some power is lost inside the source.
Series current
\[I=\frac{\mathcal E}{r+R_L}\]
Load power
\[P_L=I^2R_L=\frac{\mathcal E^2R_L}{(r+R_L)^2}\]
Efficiency
\[\eta=\frac{P_L}{P_L+I^2r}=\frac{R_L}{R_L+r}\]
Rules
Electrical power
\[P=IV\]
Power in a resistor
\[P=I^2R=\frac{V^2}{R}\]
Line loss
\[P_{\mathrm{loss}}=I^2R_{\mathrm{line}}\]
Efficiency
\[\eta=\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}\]
Maximum load power condition
\[R_L=r\]
Examples
Question
A
\[2.0\,\mathrm{kW}\]
load is supplied at \[100\,\mathrm{V}\]
Find the current.Answer
\[I=\frac{P}{V}=\frac{2000}{100}=20\,\mathrm{A}\]
Checks
- Line loss depends on current squared, so reducing current is more valuable than it first appears.
- Higher delivery voltage lowers current for the same delivered power.
- Maximum power transfer is not maximum efficiency; when \(R_L=r\), half the source power is lost internally.
- Terminal voltage falls below emf when current is drawn from a source with internal resistance.
- Use total line resistance for the complete current path, including the supply and return conductors.