High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > DC Circuits

Power and Brightness

Kirchhoff's Rules Previous Section
Real Batteries Next Section
Popular Courses
Find My Course
0:00 / 0:00

Power and Brightness





Electric circuits consume energy (and expel it in the form of heat) when current flows through them. Due to conservation of energy, all energy that is provided by the power source (battery) is consumed by the loads (resistors) in the circuit.


The power (PP) consumed by the circuit is equal to the energy used (EE), or work done (WW) by the circuit, per unit time (tt):

 P=ΔEt=Wt \boxed{ \ P=\dfrac{\Delta E}{t}=\dfrac{W}{t} \ }



The power dissipated by a resistor is directly proportional to the current (II) and voltage (VV) across it, and is given by:

 P=IV \boxed{ \ P=IV \ }



We can combine this with Ohm's law to get two other versions of the power formula:

 P=I2R=V2R \boxed{ \ P=I^2R=\dfrac{V^2}{R} \ }


Exam Tip
Use one version or the other depending on which variable is held constant:
  • If the current is constant, use P=I2RP=I^2R (e.g. for resistors in series).
  • If the voltage is constant, use P=V2RP=\dfrac{V^2}{R} (e.g. for resistors in parallel).


Wize Concept
The brightness of a light bulb is directly proportional to the power dissipated by it.

0:00 / 0:00

Example: Lightbulbs in Series and Parallel


Rank the bulbs from brightest to dimmest in the circuit below. All the bulbs have the same resistance.


First, let's look at the total current through the circuit:

I=IA+IB+ICDI=I_A+I_B+I_{CD}

The voltage drops across the parallel elements are equal:

V=VA=VB=VCDV=V_A=V_B=V_{CD}

The equivalent resistance of the two series lightbulbs will be:

RCD=RC+RD=2RR_{CD}=R_C+R_D=2R


Now we can use Ohm's Law (V=IRV=IR) to find the current through each lightbulb.

Lightbulb A:

IA=VRI_A=\dfrac{V}{R}

Lightbulb B: (same as A)

Lightbulb C:

IC=V2RI_C=\dfrac{V}{2R}

Lightbulb D: (same as C)

Note: Bulbs C and D are in series, so the current through both will be the same. But to find that current we've used the equivalent resistance (both resistors combined), and the voltage through that whole branch (which is the same as the battery voltage since the branch is in parallel with it), not the voltages through each.

Finally, we can use P=I2RP=I^2R to compare the brightness (the larger the power, the brighter the lightbulb):

Lightbulb A:

PA=IA2R=(VR)2R=V2RP_A=I_A^2R=\bigg(\dfrac{V}{R}\bigg)^2R=\dfrac{V^2}{R}

Lightbulb B: (same as A)

Lightbulb C:

PC=IC2R=(V2R)2R=V24R=14PAP_C=I_C^2R=\bigg(\dfrac{V}{2R}\bigg)^2R=\dfrac{V^2}{4R}=\dfrac{1}{4}P_A

Lightbulb D: (same as C)

Therefore the power ranking is:

PA=PB>PC=PDP_A=P_B>P_C=P_D

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: Lightbulb with Switch Open and Closed


A light bulb is connected in a circuit as shown below. Initially the switch is open. What happens to the brightness of the bulb and the potential difference across it after switch is closed?