# 9.4 Ohm's law  (Page 2/7)

 Page 2 / 7 A resistor is placed in a circuit with a battery. The voltage applied varies from −10.00 V to +10.00 V, increased by 1.00-V increments. A plot shows values of the voltage versus the current typical of what a casual experimenter might find.

In this experiment, the voltage applied across the resistor varies from −10.00 to +10.00 V, by increments of 1.00 V. The current through the resistor and the voltage across the resistor are measured. A plot is made of the voltage versus the current, and the result is approximately linear. The slope of the line is the resistance, or the voltage divided by the current. This result is known as Ohm’s law    :

$V=IR,$

where V is the voltage measured in volts across the object in question, I is the current measured through the object in amps, and R is the resistance in units of ohms. As stated previously, any device that shows a linear relationship between the voltage and the current is known as an ohmic device. A resistor is therefore an ohmic device.

## Measuring resistance

A carbon resistor at room temperature $\left(20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)$ is attached to a 9.00-V battery and the current measured through the resistor is 3.00 mA. (a) What is the resistance of the resistor measured in ohms? (b) If the temperature of the resistor is increased to $60\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ by heating the resistor, what is the current through the resistor?

## Strategy

(a) The resistance can be found using Ohm’s law. Ohm’s law states that $V=IR$ , so the resistance can be found using $R=V\text{/}I$ .

(b) First, the resistance is temperature dependent so the new resistance after the resistor has been heated can be found using $R={R}_{0}\left(1+\alpha \text{Δ}T\right)$ . The current can be found using Ohm’s law in the form $I=V\text{/}R$ .

## Solution

1. Using Ohm’s law and solving for the resistance yields the resistance at room temperature:
$R=\frac{V}{I}=\frac{9.00\phantom{\rule{0.2em}{0ex}}\text{V}}{3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{A}}=3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{Ω}\phantom{\rule{0.2em}{0ex}}=3.00\phantom{\rule{0.2em}{0ex}}\text{k}\phantom{\rule{0.2em}{0ex}}\text{Ω}.$
2. The resistance at $60\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ can be found using $R={R}_{0}\left(1+\alpha \text{Δ}T\right)$ where the temperature coefficient for carbon is $\alpha =-0.0005$ . $R={R}_{0}\left(1+\alpha \text{Δ}T\right)=3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\left(1-0.0005\left(60\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}-20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)\right)=2.94\phantom{\rule{0.2em}{0ex}}\text{k}\phantom{\rule{0.2em}{0ex}}\text{Ω}$ .
The current through the heated resistor is
$I=\frac{V}{R}=\frac{9.00\phantom{\rule{0.2em}{0ex}}\text{V}}{2.94\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{Ω}\phantom{\rule{0.2em}{0ex}}}=3.06\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\text{A}=3.06\phantom{\rule{0.2em}{0ex}}\text{mA}.$

## Significance

A change in temperature of $40\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ resulted in a 2.00% change in current. This may not seem like a very great change, but changing electrical characteristics can have a strong effect on the circuits. For this reason, many electronic appliances, such as computers, contain fans to remove the heat dissipated by components in the electric circuits.

Check Your Understanding The voltage supplied to your house varies as $V\left(t\right)={V}_{\text{max}}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\left(2\pi ft\right)$ . If a resistor is connected across this voltage, will Ohm’s law $V=IR$ still be valid?

Yes, Ohm’s law is still valid. At every point in time the current is equal to $I\left(t\right)=V\left(t\right)\text{/}R$ , so the current is also a function of time, $I\left(t\right)=\frac{{V}_{\text{max}}}{R}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\left(2\pi ft\right)$ .

See how the equation form of Ohm’s law relates to a simple circuit. Adjust the voltage and resistance, and see the current change according to Ohm’s law. The sizes of the symbols in the equation change to match the circuit diagram.

Nonohmic devices do not exhibit a linear relationship between the voltage and the current. One such device is the semiconducting circuit element known as a diode. A diode    is a circuit device that allows current flow in only one direction. A diagram of a simple circuit consisting of a battery, a diode, and a resistor is shown in [link] . Although we do not cover the theory of the diode in this section, the diode can be tested to see if it is an ohmic or a nonohmic device.

calculate ideal gas pressure of 0.300mol,v=2L T=40°c
what is principle of superposition
what are questions that are likely to come out during exam
what is electricity
watt is electricity.
electricity ka full definition with formula
Jyoti
If a point charge is released from rest in a uniform electric field will it follow a field line? Will it do so if the electric field is not uniform?
Maxwell's stress tensor is
Yes
doris
neither vector nor scalar
Anil
if 6.0×10^13 electrons are placed on a metal sphere of charge 9.0micro Coulombs, what is the net charge on the sphere
18.51micro Coulombs
ASHOK
Is it possible to find the magnetic field of a circular loop at the centre by using ampere's law?
Is it possible to find the magnetic field of a circular loop at it's centre?
yes
Brother
The density of a gas of relative molecular mass 28 at a certain temperature is 0.90 K kgmcube.The root mean square speed of the gas molecules at that temperature is 602ms.Assuming that the rate of diffusion of a gas in inversely proportional to the square root of its density,calculate the density of
A hot liquid at 80degree Celsius is added to 600g of the same liquid originally at 10 degree Celsius. when the mixture reaches 30 degree Celsius, what will be the total mass of the liquid?
Under which topic
doris
what is electrostatics
Study of charges which are at rest
himanshu
Explain Kinematics
Two equal positive charges are repelling each other. The force on the charge on the left is 3.0 Newtons. Using your notes on Coulomb's law, and the forces acting on each of the charges, what is the force on the charge on the right?
Using the same two positive charges, the left positive charge is increased so that its charge is 4 times LARGER than the charge on the right. Using your notes on Coulomb's law and changes to the charge, once the charge is increased, what is the new force of repulsion between the two positive charges?
Nya
A mass 'm' is attached to a spring oscillates every 5 second. If the mass is increased by a 5 kg, the period increases by 3 second. Find its initial mass 'm'
a hot water tank containing 50,000g of water is heated by an electric immersion heater rated at 3kilowatt,240volt, calculate the current By Madison Christian By Anh Dao By OpenStax By David Martin By OpenStax By Vanessa Soledad By Stephen Voron By Danielle Stephens By Michael Nelson By OpenStax