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Physics · Ch 11 — Electric Current Through Conductors

Electrical Energy and Power

11.7

Electrical Energy and Power

Consider a resistor ABAB connected to a cell in a circuit (Fig. 11.6), with current flowing from AA to BB. The cell maintains a potential difference VV between the two terminals of the resistor, with AA at the higher potential and BB at the lower. If a charge QQ flows through the resistor from AA to BB in time Δt\Delta t, then since VV (by definition) is the work done per unit positive charge moved from AA to BB,

V=WQ,W=VQ— (11.17)V=\frac{W}{Q},\qquad W=VQ\qquad\text{--- (11.17)}

As charge QQ moves from the higher-potential point AA to the lower-potential point BB -- i.e. through a DECREASE in potential of value VV -- its potential energy decreases by an amount

ΔU=QV=I Δt V— (11.18)\Delta U=QV=I\,\Delta t\,V\qquad\text{--- (11.18)}

By conservation of energy, this lost potential energy does not simply vanish; it is converted into some other form of energy by the resistor (or whatever device sits in place of it). Taking the limit Δt→0\Delta t\to 0,

dUdt=IV— (11.19)\frac{dU}{dt}=IV\qquad\text{--- (11.19)}

Here dU/dtdU/dt is the POWER -- the time rate at which energy is transferred -- so

P=dUdt=IV— (11.20)P=\frac{dU}{dt}=IV\qquad\text{--- (11.20)}

This is the general expression for the electrical power delivered by a cell to a resistor, or indeed to any other device (a motor, a rechargeable battery being charged, etc.) that takes its place. …

Figure 11.6A simple circuit with a cell and a resistor

What this figure shows. A simple closed circuit diagram consisting of a cell (drawn with its standard long-and-short parallel line symbol) connected by wires to a resistor labelled AB, with current shown flowing from terminal A to terminal B through the resistor. The cell maintains point A at a higher electric potential and point B at a lower electric potential; the potential difference V between A and B is marked along the resistor. No other components are shown -- this is the minimal circuit used to derive that the work done carrying charge Q from A to B is W = VQ, and that the power del …

Misc Ex.4Example 11.4 -- Power and energy consumed by an electric heater

Worked out. An electric heater takes 6 A current from a 230 V supply line; the worked solution first computes the power P = IV = 6×230 = 1380 W = 1.38 kW, then multiplies by the given time of 5 hours to get the energy consumed, Energy = Power × time = 1.38 kW × 5 h = 6.90 kWh, and states this equivalently as 6.9 units of electrical energy (since 1 kWh is defined as 1 'unit' of electrical energy for billing purposes) …