4.4Electrical power

Since we often use electricity to convey energy from one location to another, it is important to be able to quantify and calculate the rate at which energy is delivered by an electric circuit. The rate at which energy transfers from one location to another is technically referred to as power (P ), and it is typically4 measured in the unit of the watt (1 watt = 1 joule of energy per second of time).

The total amount of power (energy per unit time) carried by an electric current is proportional to the strength of that current as well as the amount of potential (voltage) between the poles of the electrical source. Thus, power is equal to the product of current and voltage in a formula called

Joules’ Law :

P = IV

Where,

P = Power in watts (joules of energy per second of time)

I = Current in amperes

V = Voltage in volts

We may prove the correctness of this formula by verifying all the units of measurement agree:

Note how the basic units for power (joules per second) are indeed equal to the product of voltage (joules per coulomb) and current (coulombs per second). This process of checking for agreement between units of measurement in a physics formula is called dimensional analysis.

4Although the unit of the “watt” is commonly used for electrical power, other units are valid as well. The British unit of horsepower is every bit as valid for expressing electrical power as “watts,” although this usage is less common. Likewise, the “watt” may be used to express measurements of non-electrical power as well, such as the mechanical power output of an engine. European automobile manufacturers, for example, rate the power output of their cars’ engines in kilowatts, as opposed to American automobile manufacturers who rate their engines in horsepower. This choice of units is strictly a cultural convention, since any valid unit for power may be applied to any form of energy rate.

If the circuit in question has only one source of voltage, one resistance, and one path for current, we cannot mis-apply Ohm’s Law. Expressing the previous example in a schematic diagram:

Current

source

wire

7.143 mA

Current

Note: arrows point in the direction of conventional flow

However, if we look at a more complex circuit, we encounter the potential for mis-application of Ohm’s Law because we have multiple resistances in the circuit rather than just one resistance:

+

R1 3500 Ω

25 V −

R2 1500 Ω

Which resistance do we use to calculate current in this circuit? Do we divide our 25 volts by 3500 ohms like we did last time, or do we divide it by 1500 ohms, or something entirely di erent? The answer to this question lies in the identification of voltages and currents. We know that the 25 volt potential will be impressed across the total of the two resistances R1 and R2, and since there is only one path for current they must share the same current. Thus, we actually have three voltages (V1, V2, and Vtotal), three resistances (R1, R2, and Rtotal), and only one current (I):

I

I

Note: arrows point in the direction of conventional flow

Using the V = IR form of Ohm’s Law to relate these three voltages (V1, V2, and Vtotal) to the one current (I), we end up with three equations for this circuit:

V1 = IR1

V2 = IR2

Vtotal = IRtotal = I(R1 + R2)

We can only solve for one unknown variable at a time in any equation. This means we cannot solve for V1 yet, because although we do know the value of resistor R1 (3500 ohms), we do not yet know the circuit’s current (I). Likewise for V2, because we do not yet know the value of I. However, the third equation is solvable, since we happen to know the total voltage as well as both resistor values, leaving circuit current I as the only unknown variable. Manipulating this equation and solving for I:

Now that we know the amount of current in this circuit, we may solve for V1 and V2 in the other equations to find that the voltage across resistor R1 is 17.5 volts, and that the voltage across resistor R2 is 7.5 volts.