Ohm's Law Example Problems With Solutions for High School

Experimentally found that when a voltage or potential difference $\Delta V$ is applied across the ends of certain conductors, the current through them is proportional to the applied voltage, that is $I \propto \Delta V$.

The proportionality constant is called the resistance of that conductor.

In other words, resistance is defined as the ratio of the voltage across a conductor to the current flowing through it.\[R \equiv \frac{\Delta V}{I}\] This simple relationship between potential difference and current is known as Ohm's law

The SI units of resistance are volts per ampere which are called Ohms ($\Omega$).

A conductor that implements a specific amount of resistance in an electrical circuit is called the resistor.

For example, if a 10 ohms resistor is connected to the terminals of a battery with voltage 240 volts, then a current of $\frac{240}{10}=24\,{\rm A}$ passes through it.

In contrast, there are also conductors or materials in electronics that above simple linear relationship between voltage and current is not held such as diodes, transistors, or fluorescent lamps.

In such materials, there is a nonlinear voltage-current relationship. These conductors are called non-ohmic materials.

Are you getting ready for the AP Physics exam? Read this:
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In the following, some simple questions and answers about Ohm's law are provided with detailed explanations. All problems are suitable for a high school student.

Ohm's Law Examples

Example (1): An electronic device has a resistance of 20 ohms and a current of 15 A. What is the voltage across the device?

Solution: resistance, current, and voltage are related together by Ohm's law as $V=IR$. Thus, the voltage of the device is obtained as \begin{align*}V&=IR\\&=15\times 20\\&=300\quad {\rm V}\end{align*}


Example (2): a $3-{\rm V}$ potential difference is applied across a $6\,{\rm \Omega}$ resistor. What is the current that flows into the resistor?

Solution: Ohm's law states the potential difference across a resistor is resistance times current so we get \begin{align*} I&=\frac VR\\&=\frac {3}{6}\\&=0.5\quad {\rm A}\end{align*}

Homework: In an experiment of measuring the current through an unknown resistor, a student obtained the following data. 

Voltage (V) Current (I)
3.0 0.151
6.0 0.310
9.0 0.448
12.0 0.511
15.0 0.750

(a) By drawing a graph on paper, show the relationship between current and voltage.
(b) Using that graph, determine the resistance of the resistor.

The solution to this and 34 other homework problems are here.


Example (3): A current of $0.2\,{\rm A}$ passes through a $1.4\,{\rm k \Omega}$ resistor. What is the voltage across it?

Solution: using Ohm's law, $V=I R$, we have \begin{align*}V&=IR\\&=(0.2\,{\rm A})(1.4\times 1000\,{\rm  \Omega})\\&=280\quad {\rm V}\end{align*}

Example (4): In the circuit shown below, how much current does the ammeter show?

Ammeter-lamp in a series circuit

Solution: the lamp is an electronic component with high resistance. In the figure, the voltage across it is the same as battery $V=20\,{\rm V}$. The current passes through it relates to the resistance and voltage drop using Ohm's law \begin{align*} I&=\frac VR\\&=\frac{20}{8}\\&=1.25\quad {\rm A}\end{align*}

Example (5): In a circuit, the potential drop across the 10 kΩ -resistor is 100 V. What is the current through the resistor? 

Solution: Substitute all known numerical values into the Ohm's equation, $V=IR$. \begin{align*} I&=\frac VR\\\\&=\frac{100\,{\rm V}}{10000\,{\rm \Omega}}\\\\&=0.01\quad {\rm A}\end{align*}


Example (6): in the following circuits, find the unknowns. electric circuit with resistance and lamp

Solution: In each of the circuits, use Ohm's law $V=IR$ and solve for the unknown. In the left circuit, the current through the resistor is asked in milliamps. Thus, \begin{align*} I&=\frac VR\\\\&=\frac{120}{100}\\\\&=1.2\quad {\rm A}\end{align*} To convert it to milliamperes, multiply it by 1000 so we get $I=1200\,{\rm mA}$.

In the right circuit, the resistance of the light bulb is unknown. Thus, \begin{align*} R&=\frac VI\\&=\frac {24}{600\times 10^{-3}} \\\\&=40\quad {\rm \Omega}\end{align*}

Example (7): The voltage-current curve for an ohmic conductor is plotted as shown in the figure below. What is the resistance of resistors 1 and 2?voltage-current for ohmic conductor

Solution: Ohm's law tells us that resistance is the slope of the voltage vs. current curve $R =\frac{\Delta V}{I}$. Recall that the slope $m$ of a straight line between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is determined as \[m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\] Thus, the slope of the voltage - current curve , which is the resistance, is obtained as below:
The points $A(0,0)$ and $B(2,20)$ are on the line (1):\[R_1=\frac{20-0}{2-0}=10\quad{\rm \Omega}\]
The points $A(0,0)$ and $B(4,10)$ are on the line (2):\[R_2=\frac{10-0}{4-0}=2.5\quad{\rm \Omega}\]

Example (8): Reverse the position of the potential drop across a conductor and the current passing through it in the previous problem, so that the current-voltage curve is obtained. Now, find the resistance of resistors 1 and 2?

Solution: if we rearrange Ohm's law as $I=\frac{1}{R}\Delta V$, we can see that the slope of the current- voltage curve, in this case, gives the inverse of the resistance. Therefore, as in the previous problem, 
slope of the line (1) is \[\frac{1}{R_1}=\frac{20-0}{2-0}=10\] which get the $R_1=0.1\,{\rm \Omega}$ and slope of the line (2) is \[\frac{1}{R_2}=\frac{10-0}{4-0}=2.5\] which gives $R_2=0.4\,{\rm \Omega}$. 


Example (9): A student conducts an experiment and measures the current and voltage across two unknown resistors. Then she plots her finding in a current-voltage-current curve for nonohmic materialvoltage coordinate, as shown in the figure. What can be said about resistors A and B? 

Solution: ohmic materials are the ones that have constant resistance over a wide range of applied voltages. In other words, in an ohmic conductor, the ratio of the voltage across it to the current through it, which is defined as resistance, is always a constant. 

Thus, ohmic materials have a linear current-voltage relationship, and its curve passes through the origin. In contrast, the materials having a resistance that changes with a potential drop or current are called non-ohmic. 

The curve of a non-ohmic material is not linear. Examples of non-ohmic materials that violate Ohm's law are diodes and transistors. 

With these explanations, as the curve of (A) is linear, and passes through the origin, so it is an ohmic conductor whose slope gives the reciprocal of the resistance. As in the previous problem, its resistance is calculated as $R_A=5\,{\rm \Omega}$. 

The resistor (B) has a nonlinear relationship between the voltage across it to the current, so it is a non-ohmic conductor with variable resistance. 

Ohm's Law: Practice Problems with Solution 

Now, We want to solve some practice problems to show you how to use ohm's law to solve electricity problems.

Practice Problem (1): An alarm clock draws 0.5 A current when connected to a 120 V circuit. Find its resistance. 

Solution: current $I=0.5\,{\rm A}$ and voltage drop $V=120\,{\rm V}$ given. Solve Ohm's law for unknown $R$ as \begin{align*} R&=\frac VI\\ \\&=\frac{120}{0.5}\\ \\&=240\quad {\rm \Omega}\end{align*}

Practice Problem (2): A sub-woofer needs a household voltage of 110 V to push a current of 5.5 A through its coil. What is the resistance of the sub-woofer?

Solution: Known are voltage difference $V=110\,{\rm V}$ and current $I=5.5\,{\rm A}$. Ohm's law relates them as below \begin{align*} R&=\frac VI \\ \\ &=\frac{110}{5.5} \\ \\ &=20\quad {\rm \Omega}\end{align*}


Problem (3): How much current is drawn from a circuit that has a resistor of 1000 ohms if powered by a battery with 1.5 volts.

Solution: Resistance $R=1000\,{\rm \Omega}$ and voltage $V=1.5\,{\rm V}$ are known, so we have \begin{align*}I&=\frac VR\\\\&= \frac{1.5}{1000} \\\\&= 1.5\quad {\rm mA}\end{align*}

Problem (4): An electric heater has a coiled metal wire which draws a current of 100 A. The resistance of the wire is 1.1 ohms. Calculate the voltage that must be established to it. 

Solution: Current $I=100\,{\rm A}$ and resistance $R=1.1\,{\rm \Omega}$ are related as \begin{align*} V&=IR\\&=100\times 1.1\\&=110\quad {\rm V}\end{align*}

Problem (5): Maximum current which passes through a light bulb with a resistance of 5 ohms is 10 A. How much voltage must be applied across its ends before the bulb will break?

Solution: Maximum voltage can be found using Ohm's law as below \begin{align*} V&=IR \\ &= 10\times 5\\&=50 \quad {\rm V}\end{align*} If a voltage higher than this value is applied on the circuit, the lamp will burn out.  

Problem (6): In a circuit, we replace the previous 1.5-volt battery with a 3-volt new one. What happens to this circuit? 

Solution: Ohm's law tells us that when more voltage establishes across a circuit, then a higher current would flow through the resistors in a circuit like electric heaters, light bulbs, and so on. 

More current can cause damage or failure to the household appliances. For example, a light bulb with resistance $R=1.5\,{\rm \Omega}$ draws a current of $I=\frac{1.5}{1.5}=1\,{\rm A}$ with a $1.5$ volts battery and a current $I=\frac{3}{1.5}=2\,{\rm A}$ with replacement new one. In these cases, the light bulb will most likely burn out. 

Problem (7): In a circuit, a $10\,{\rm \Omega}$ resistor is removed and replaced by a $20\,{\rm \Omega}$ resistor. What happens to the current in the circuit. 

Solution: Since nothing said about the voltage drop across the circuit, we assume it is constant say, $V=120\,{\rm V}$. Therefore, using Ohm's law formula, $I=\frac VR$, current $I=\frac{120}{10}=12\,{\rm A}$ flows through the $10\,{\rm \Omega}$ resistor and $I=\frac{120}{20}=6\,{\rm A}$ through the $20\,{\rm \Omega}$ resistor. 

We can see that for the same voltage, doubling the resistances results in decreasing, more precisely halving the currents. 

In this article, we learned how to solve problems involving Ohm's law with many solved examples. 


Author: Dr. Ali Nemati
Page Created: 12/6/2020
Last Update: 1/19/2021