<meta http-equiv="refresh" content="1; url=/nojavascript/"> Ohm's Law ( Read ) | Physics | CK-12 Foundation
You are viewing an older version of this Concept. Go to the latest version.

# Ohm's Law

%
Progress
Practice Ohm's Law
Progress
%
Ohm's Law

Students will learn a basic understanding of electric circuits and how to apply Ohm's law.

### Key Equations

Ohm’s Law

$V=IR \;$ ; Voltage drop equals current multiplied by resistance.

Power

$P = IV \;$ ; Power released is equal to the voltage multiplied by the current.

Guidance
• Ohm's Law is the main equation for electric circuits but it is often misused. In order to calculate the voltage drop across a light bulb use the formula: $V_{lightbulb} = I_{lightbulb}R_{lightbulb}$ . For the total current flowing out of the power source, you need the total resistance of the circuit and the total voltage: $V_{total} = I_{total}R_{total}$ .
• The equations used to calculate the power dissipated in a circuit is $P=IV$ . As with Ohm’s Law, one must be careful not to mix apples with oranges. If you want the power of the entire circuit, then you multiply the total voltage of the power source by the total current coming out of the power source. If you want the power dissipated (i.e. released) by a light bulb, then you multiply the voltage drop across the light bulb by the current going through that light bulb.
• Power is the rate that energy is released. The units for power are Watts $(W)$ , which equal Joules per second $[W] = [J]/[s]$ . Therefore, a $60\;\mathrm{W}$ light bulb releases $60$ Joules of energy every second.

#### Example 2

A small flash light uses a single AA battery which provides a voltage of 1.5 V. If the bulb has a resistance of $2\;\Omega$ , how much power is dissipated by the light bulb.

##### Solution

Since the light bulb is the only object in the circuit, we know the voltage drop across the light bulb is equal to that of the battery. Therefore, we can use Ohm's law to solve for the current in the resistor.

$V&=IR\\I&=\frac{V}{R}\\I&=\frac{1.5\:\text{V}}{2\:\Omega}\\I&=.75\:\text{A}\\$

Now we can determine the power of the bulb.

$P&=IV\\P&=.75\:\text{A}*1.5\:\text{V}\\P&=1.13\:\text{W}\\$

### Time for Practice

1. A light bulb with resistance of $80\ \Omega$ is connected to a $9\;\mathrm{V}$ battery.
1. What is the electric current going through it?
2. What is the power (i.e. wattage) dissipated in this light bulb with the $9\;\mathrm{V}$ battery?
3. How many electrons leave the battery every hour?
4. How many Joules of energy leave the battery every hour?

2. 2. A bird is standing on an electric transmission line carrying $3000\;\mathrm{A}$ of current. A wire like this has about $3.0 \times 10^{-5}\ \Omega$ of resistance per meter. The bird’s feet are $6\;\mathrm{cm}$ apart. The bird, itself, has a resistance of about $4 \times 10^5\ \Omega.$
1. What voltage does the bird feel?
2. What current goes through the bird?
3. What is the power dissipated by the bird?
4. By how many Joules of energy does the bird heat up every hour?
3. 3. A $120\;\mathrm{V}, 75\;\mathrm{W}$ light bulb is shining in your room and you ask yourself…
1. What is the resistance of the light bulb?
2. How bright would it shine with a $9\;\mathrm{V}$ battery (i.e. what is its power output)?
4. 4. Students measure an unknown resistor and list their results in the Table ( below ); based on their results, complete the following:
1. Show a circuit diagram with the connections to the power supply, ammeter and voltmeter.
2. Graph voltage vs. current; find the best-fit straight line.
3. Use this line to determine the resistance.
4. How confident can you be of the results?
5. Use the graph to determine the current if the voltage were $13\;\mathrm{V}$ .
Voltage $(v)$ Current $(a)$
$15$ $.11$
$12$ $.08$
$10$ $.068$
$8$ $.052$
$6$ $.04$
$4$ $.025$
$2$ $.01$
5. 5. A certain 48-V electric forklift can lift up to 7000 lb at a maximum rate of 76 ft/min.
1. What is its power?
2. What current must the battery produce to achieve this power?

1. a. $0.11 \;\mathrm{A}$ b. $1.0 \;\mathrm{W}$ c. $2.5 \times 10^{21}$ electrons d. $3636 \;\mathrm{W}$
2. a. $5.4 \;\mathrm{mV}$ b. $1.4 \times 10^{-8} \;\mathrm{A}$ c. $7.3 \times 10^{-11} \;\mathrm{W}$ , not a lot d. $2.6 \times 10^{-7} \;\mathrm{J}$
3. a. $192 \ \Omega$ b. $0.42 \;\mathrm{W}$
4. .
5. a. 12300 W b. 256 A