3.4 - Electric Power

Power in Circuits

Power is how fast energy gets transferred, converted, or dissipated by a circuit element.

How much power an element handles depends on the current running through it and the potential difference across it:

P = IV

where: -

P

= power (watts, W) -

I

= current through the element (A) -

V

= potential difference across the element (V)

Unit: 1 watt (W) is just 1 joule per second (J/s).

The equation sheet writes this as

P = I\Delta V

, with

\Delta V

for the potential difference across the element. The next two forms are derived equations: they aren’t on the sheet, so be ready to build them from

P = I\Delta V

and

I = \frac{\Delta V}{R}

Derived Power Equations

If you plug in Ohm’s law (

V = IR

), you can rewrite the power equation in a couple of other forms.

Power in terms of current and resistance

P = I^2R

Where it comes from:

substitute

V = IR

into

P = IV

P = I(IR) = I^2R

Power in terms of voltage and resistance

P = \frac{V^2}{R}

Where it comes from:

substitute

I = \frac{V}{R}

into

P = IV

P = \frac{V}{R} \cdot V = \frac{V^2}{R}

When to Use Each Formula

All three versions give you the same answer, so the right one to reach for just depends on what you already know:

$P = IV$ works when you have current and voltage.

$P = I^2R$ works when you have current and resistance, or when you’re comparing elements that share the same current.

$P = \frac{V^2}{R}$ works when you have voltage and resistance, or when you’re comparing elements that share the same voltage.

Power and Energy

Since power is just the rate at which energy moves:

P = \frac{E}{\Delta t}

which rearranges to:

E = P \cdot \Delta t

where

E

is the energy transferred (in joules) over a time interval

\Delta t

Energy in, energy out. A battery transfers energy into the circuit at a rate

P = I\Delta V

, where

\Delta V = \mathcal{E}

for an ideal battery. Resistors and bulbs transfer energy out of the circuit (as thermal energy and light), each at its own rate

P = I\Delta V

. Energy is conserved, so the battery’s power output equals the total power used by everything else in the circuit.

Example: a 12 V battery is connected to a 4.0 Ω resistor and a 2.0 Ω resistor in series. Compare the power the battery supplies with the power the resistors use.

I = \frac{\mathcal{E}}{R_1 + R_2} = \frac{12 \text{ V}}{4.0\ \Omega + 2.0\ \Omega} = 2.0 \text{ A}

P_{\text{battery}} = I\mathcal{E} = (2.0 \text{ A})(12 \text{ V}) = 24 \text{ W}

P_1 = I^2R_1 = (2.0 \text{ A})^2(4.0\ \Omega) = 16 \text{ W}, \qquad P_2 = I^2R_2 = (2.0 \text{ A})^2(2.0\ \Omega) = 8.0 \text{ W}

Answer: the battery supplies 24 W, and the resistors use

16 + 8.0 = 24

W. Every joule the battery puts in comes out as thermal energy in the resistors.

Power and Bulb Brightness

A bulb gets brighter as its power goes up. The more power a bulb dissipates, the brighter it glows.

That’s handy because it means you can predict and compare bulb brightness in a circuit just by working out or comparing how much power each one dissipates.

Example (ranking brightness): three identical bulbs, each 6.0 Ω, are connected to a 12 V battery. Bulb A is in series with the battery, and bulbs B and C are in parallel with each other after it. Rank the bulbs by brightness.

B and C in parallel act like one 3.0 Ω resistor (

\frac{1}{R} = \frac{1}{6.0} + \frac{1}{6.0}

), so the whole circuit is

6.0 + 3.0 = 9.0

Ω. All the current goes through A (keeping an extra digit until the end):

I_A = \frac{12 \text{ V}}{9.0\ \Omega} = 1.33 \text{ A}

That current splits evenly between the identical bulbs B and C, so

I_B = I_C = 0.667

P_A = I_A^2R = (1.33 \text{ A})^2(6.0\ \Omega) \approx 11 \text{ W}, \qquad P_B = P_C = (0.667 \text{ A})^2(6.0\ \Omega) \approx 2.7 \text{ W}

Answer: A is the brightest, and B and C are tied and dimmer. A carries twice the current of B, and power goes as

I^2

, so A dissipates 4 times as much power.

Follow-up: if bulb C is unscrewed, its branch becomes an open circuit, and A and B are simply in series:

I = \frac{12 \text{ V}}{12\ \Omega} = 1.0

A, so

P_A = P_B = (1.0 \text{ A})^2(6.0\ \Omega) = 6.0

W. Bulb A gets dimmer (about 11 W down to 6.0 W) and bulb B gets brighter (about 2.7 W up to 6.0 W).

Practice Problems

1. A resistor has 9.0 V across it and carries a current of 1.5 A. - (a) How much power does the resistor dissipate? - (b) How much energy turns into thermal energy in 2.0 minutes?

2. Two resistors,

R_1 = 10

Ω and

R_2 = 20

Ω, are wired in series with a battery. If the current through the circuit is 0.50 A, which resistor dissipates more power, and by how much?

3. A 60 W lightbulb runs at 120 V. - (a) What’s the current through the bulb? - (b) What’s the resistance of the bulb’s filament?

4. Two identical bulbs are hooked up to a battery. In circuit A they’re in series, and in circuit B they’re in parallel. In which circuit do the bulbs shine brighter? Explain it using power.

5. A heating element has a resistance of 15 Ω. If it draws 8.0 A, how much power does it use?

Solutions

Use the basic power equation:

P = IV = (1.5 \text{ A})(9.0 \text{ V}) = 13.5 \text{ W}

Answer: it dissipates 13.5 W.

First convert the time to seconds:

\Delta t = 2.0 \text{ min} \times \frac{60 \text{ s}}{1 \text{ min}} = 120 \text{ s}

Then find the energy:

E = P \cdot \Delta t = (13.5 \text{ W})(120 \text{ s}) = 1620 \text{ J} = 1.62 \text{ kJ}

Answer: 1620 J (or 1.62 kJ) becomes thermal energy.

2. They’re in series, so they carry the same current. That makes

P = I^2R

the natural choice.

For

R_1

P_1 = I^2R_1 = (0.50 \text{ A})^2(10 \text{ Ω}) = 2.5 \text{ W}

For

R_2

P_2 = I^2R_2 = (0.50 \text{ A})^2(20 \text{ Ω}) = 5.0 \text{ W}

Comparing them:

\frac{P_2}{P_1} = \frac{5.0 \text{ W}}{2.5 \text{ W}} = 2

Answer:

R_2

dissipates twice as much power. In series, the bigger resistor dissipates more.

Rearrange $P = IV$ to get current:

I = \frac{P}{V} = \frac{60 \text{ W}}{120 \text{ V}} = 0.50 \text{ A}

Answer: the current is 0.50 A.

Use Ohm’s law or the power equation $P = \frac{V^2}{R}$ :

R = \frac{V^2}{P} = \frac{(120 \text{ V})^2}{60 \text{ W}} = \frac{14400}{60} = 240 \text{ Ω}

Answer: the resistance is 240 Ω.

4. Circuit B, the parallel one, has the brighter bulbs.

In the series circuit, the same current runs through both bulbs. Total resistance is higher because you’re adding the two resistances together, so for a given battery voltage the current ends up lower. Each bulb gets the full current, but that current is small. Power per bulb here is

P_{\text{series}} = I^2R_{\text{bulb}}

In the parallel circuit, each bulb sits directly across the battery terminals, so each one gets the full battery voltage. Using

P = \frac{V^2}{R}

, each bulb dissipates more power because it has the full voltage across it. On top of that, the total current pulled from the battery is higher.

Since each parallel bulb dissipates more power than each series bulb, the parallel bulbs win on brightness.

5. You know current and resistance, so use

P = I^2R

P = I^2R = (8.0 \text{ A})^2(15 \text{ Ω}) = (64)(15) = 960 \text{ W}

Answer: the heating element uses 960 W.

Common Mistakes

Grabbing the wrong power formula. Pick based on what you’re given and what stays constant in the circuit. In series the current is shared (use $P = I^2R$ ), and in parallel the voltage is shared (use $P = \frac{V^2}{R}$ ).

Mixing up power and energy. Power is the rate of energy transfer (J/s), while energy is the total amount transferred (J). If you want energy, don’t forget to multiply by time.

Assuming more resistance always means more power. It depends on the setup. In series (constant $I$ ), higher $R$ means more power ( $P = I^2R$ ). In parallel (constant $V$ ), higher $R$ means less power ( $P = \frac{V^2}{R}$ ).

Forgetting to square or divide. In $P = I^2R$ the current gets squared. In $P = \frac{V^2}{R}$ the voltage gets squared and then divided by $R$ .

Unit slip-ups. Power should come out in watts (W), where 1 W = 1 J/s. When you’re finding energy, make sure time is in seconds.

Botching brightness comparisons. A bulb’s brightness tracks the power it dissipates, not the current through it or the voltage across it on their own.

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