Category : Electric Circuits

All topics about electric circuits including the definition of resistance and capacitance along with many solved example problems are covered.

Ohm's Law Example Problems With Solutions for High School

# 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, the resistance is defined as the ratio of the voltage across a conductor to the current 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 $10-\Omega$ resistor is connected to the terminals of a battery with voltage $240\,{\rm V}$, 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 lamp. In such materials, there is a nonlinear voltage-current relationship. These conductors are called nonohmic materials.

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.

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\\&=20\times 15\\&=300\quad {\rm V}\end{align*}

Example (2): a 3-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 the current so we get \begin{align*} I&=\frac VR\\&=\frac {3}{6}\\&=0.5\quad {\rm A}\end{align*}

Example (3): A current of 0.2 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\,{\rm k \Omega})\\&=280\quad {\rm V}\end{align*}

Example (4): In the circuit shown below, what is the resistance of the lamp?

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 the figure below, the potential drop across the 10 kΩ -resistor is 100 V. What is the current through the resistor?

Solution: plug all known 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.

Solution: In each of the circuits, use Ohm's law $V=IR$ and solve for the unknown. In the right circuit, the current through the resistor asked in milliamperes. Thus, \begin{align*} I&=\frac VR\\&=\frac{120}{100}\\&=1.2\quad {\rm A}\end{align*} To convert it to milliamperes, divide it by 1000 so we get $I=0.0012\,{\rm mA}$.

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?

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 of a straight $m$ 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 curve voltage - current, 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 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 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 nonohmic.

The curve of a nonohmic material is not linear. Examples of nonohmic materials that violate Ohm's law are diode and transistor.

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 nonohmic conductor with variable resistance.

Page Created: 12/6/2020

Author: Ali Nemati