current work power and source efficiency

Learning objectives

Outcome
Calculate the work of current in a circuit section using $A = UIt$ and $A = I^2Rt$ for given current, voltage and time.
Outcome
Distinguish the total source power $P_{\text{tot}} = \varepsilon I$ , the useful power $P_{\text{use}} = I^2R$ and the losses $P_{\text{loss}} = I^2r$ , computing each of them.
Outcome
Determine the source efficiency with $\eta = \dfrac{R}{R+r}$ and find the load resistance at which the useful power is maximum.

Understand the idea

Work and Power of Current and Source Efficiency

Text

In a complete circuit, a source with EMF $\varepsilon$ and internal resistance $r$ is connected to an external section of resistance $R$ .

Ohm's law for the complete circuit gives the current $I = \dfrac{\varepsilon}{R+r}$ , and the voltage across the external section is $U = IR = \varepsilon - Ir$ .

The work of the current in a section at voltage $U$ is $A = UIt$ ; for a resistor the same relations take the form $A = I^2Rt = \dfrac{U^2}{R}t$ , and the power is $P = UI = I^2R = \dfrac{U^2}{R}$ .

For a source we distinguish three powers:

total (supplied) power $P_{\text{tot}} = \varepsilon I$ ;
useful power (in the external circuit) $P_{\text{use}} = UI = I^2R$ ;
losses inside the source $P_{\text{loss}} = I^2r$ .

The efficiency of the source is the ratio of the useful power to the total power:

\eta = \frac{P_{\text{use}}}{P_{\text{tot}}} = \frac{U}{\varepsilon} = \frac{R}{R+r}.

As $R \to \infty$ the efficiency tends to one, but the current and the useful power tend to zero; the useful power is maximum at $R = r$ , where $\eta = 0{,}5$ .

Worked examples

Direct

Problem

A source has EMF $\varepsilon = 12$ V and internal resistance $r = 1$ Ω; the external resistance is $R = 5$ Ω. Find the current, the useful power, the total power and the efficiency of the source.

Solution

The current follows from Ohm's law for the complete circuit:

I = \frac{\varepsilon}{R+r} = \frac{12}{6} = 2\ \text{A}.

The useful power is released in the external resistor $R$ :

P_{\text{use}} = I^2R = 2^2 \cdot 5 = 20\ \text{W}.

The total power of the source:

P_{\text{tot}} = \varepsilon I = 12 \cdot 2 = 24\ \text{W}.

Check: the losses inside the source are $P_{\text{loss}} = I^2r = 4$ W, and $20 + 4 = 24$ W.

Efficiency:

\eta = \frac{P_{\text{use}}}{P_{\text{tot}}} = \frac{20}{24} \approx 0{,}83 \ (83\,\%).

Application

Problem

An electric heater of resistance $R = 10$ Ω is connected to a battery with EMF $\varepsilon = 12$ V and internal resistance $r = 2$ Ω and operates for $t = 5$ min. Find the work of the current in the heater, the energy released inside the source, and the efficiency of this circuit.

Solution

Convert the time to seconds: $t = 5 \cdot 60 = 300$ s. The current in the circuit:

I = \frac{\varepsilon}{R+r} = \frac{12}{12} = 1\ \text{A}.

The work of the current in the heater (useful energy):

A_{\text{use}} = I^2Rt = 1^2 \cdot 10 \cdot 300 = 3000\ \text{J} = 3\ \text{kJ}.

The energy released inside the source:

A_{\text{loss}} = I^2rt = 1^2 \cdot 2 \cdot 300 = 600\ \text{J}.

Check via the total work: $A_{\text{tot}} = \varepsilon I t = 12 \cdot 1 \cdot 300 = 3600$ J, and $3000 + 600 = 3600$ J.

Efficiency:

\eta = \frac{R}{R+r} = \frac{10}{12} \approx 0{,}83 \ (83\,\%).

Transfer

Problem

A source has EMF $\varepsilon$ and internal resistance $r$ , and the external resistance $R$ can be varied. Prove that the useful power is maximum at $R = r$ , and find that maximum value and the source efficiency in this regime.

Solution

The useful power as a function of the load:

P(R) = I^2R = \frac{\varepsilon^2 R}{(R+r)^2}.

Study it for a maximum by setting the derivative to zero:

\frac{dP}{dR} = \varepsilon^2 \cdot \frac{(R+r)^2 - R \cdot 2(R+r)}{(R+r)^4} = \varepsilon^2\,\frac{r - R}{(R+r)^3}.

The derivative vanishes at $R = r$ . For $R < r$ it is positive, for $R > r$ it is negative, so this is a maximum.

Substitute $R = r$ :

P_{\max} = \frac{\varepsilon^2 r}{(2r)^2} = \frac{\varepsilon^2}{4r}.

The efficiency in this regime:

\eta = \frac{R}{R+r} = \frac{r}{2r} = 0{,}5.

Thus the useful power is maximum when the external and internal resistances are equal, but then exactly half of the power is lost inside the source.

Practice

Text

Carry out the measurement in a simulation (for example, PhET "Circuit Construction Kit" or a similar one). Build the circuit: a source with $\varepsilon = 12$ V and internal resistance $r = 1$ Ω, in series with it an ammeter and a variable resistor $R$ , and in parallel with the resistor a voltmeter.

Set $R$ : 0.2; 0.5; 1; 2; 4; 8 Ω. For each value record the current $I$ and the voltage $U$ .
Compute $P_{\text{use}} = UI$ , $P_{\text{tot}} = \varepsilon I$ and $\eta = U/\varepsilon$ .
Plot $P_{\text{use}}(R)$ and $\eta(R)$ and check that the maximum useful power occurs at $R \approx r$ .

Work only with the low-voltage simulation; do not assemble this circuit with mains voltage.

Visual reference

Purpose

Show that the useful power and the source efficiency depend on the ratio $R/r$ in different ways, and that their maxima do not coincide.

Representation

Graph: the horizontal axis is the ratio $R/r$ (from 0 to 10), the vertical axis shows normalized quantities. The useful power curve $P_{\text{use}}/(\varepsilon^2/4r)$ passes through a maximum of $1$ at $R/r = 1$ and falls to zero on both sides. The efficiency curve $\eta = \dfrac{R/r}{R/r + 1}$ increases monotonically from $0$ to $1$ and equals $0{,}5$ at $R/r = 1$ . The regime of maximum useful power is marked by the vertical dashed line $R = r$ .

Real-world examples

Fast charging a smartphone

Scenario

A charger feeds a phone battery with a voltage of $5\ \text{V}$ at a current of $3\ \text{A}$ for $30\ \text{min}$ . During this time the source does the work $W = UIt = 5 \cdot 3 \cdot 1800 = 27\,000\ \text{J} = 27\ \text{kJ}$ , while the charge delivered to the battery is $q = It = 3 \cdot 0.5 = 1.5\ \text{A·h}$ — about a third of the $4000\ \text{mA·h}$ battery capacity. The phone body noticeably warms up in the process.

Connection

The power formula $P = UI = 15\ \text{W}$ and the work formula $W = UIt$ apply here; the fraction of energy that goes into heating the case and the internal resistance instead of driving the chemical charging reaction is exactly what sets the efficiency of the charging process.

An electric kettle

Scenario

A kettle rated $2000\ \text{W}$ at $220\ \text{V}$ heats $1.5\ \text{L}$ of water from $20$ to $100\ ^\circ\text{C}$ . This needs $Q = cm\Delta T = 4200 \cdot 1.5 \cdot 80 = 504\,000\ \text{J}$ . At full power it would take $t = Q/P \approx 250\ \text{s}$ , but a real kettle boils in about $300\ \text{s}$ .

Connection

The formulas $P = UI$ (power of the current) and $W = Pt$ (work of the current) are used. The gap between the predicted $250\ \text{s}$ and the observed $300\ \text{s}$ is heat lost through the walls and lid, which makes the kettle's efficiency less than one: $\eta \approx 250/300 \approx 83\ \%$ .

A car starter motor

Scenario

A car battery has an emf $\varepsilon = 12\ \text{V}$ and an internal resistance $r = 0.05\ \Omega$ . The starter draws a current $I = 100\ \text{A}$ . The terminal voltage sags to $U = \varepsilon - Ir = 12 - 100 \cdot 0.05 = 7\ \text{V}$ . The useful power is $UI = 700\ \text{W}$ , the loss inside the battery is $I^2 r = 500\ \text{W}$ , the total power of the source is $\varepsilon I = 1200\ \text{W}$ , and the efficiency is $\eta = U/\varepsilon = 7/12 \approx 58\ \%$ .

Connection

This applies the complete-circuit Ohm law $\varepsilon = I(R + r)$ directly: the source efficiency $\eta = U/\varepsilon = R/(R+r)$ drops when the external resistance $R = 0.07\ \Omega$ is comparable to the internal one, and the lost power $I^2 r$ heats the battery.