Ohm's Law and Resistive Circuits

Three things to know

Think of water in a pipe. A pump pushes the water. The pipe lets only some water through. In a circuit, it is the same.

• Current $I$ is the flow of charge. Its unit is the ampere (A).
• Voltage $V$ is the push, the potential difference across a part of the circuit. Its unit is the volt (V).
• Resistance $R$ tells how hard it is for the current to pass. Its unit is the ohm ($\Omega$).

Ohm's law

For many conductors, if the temperature stays the same, the current is proportional to the voltage across them. This is Ohm's law.

Let us see what it says.

• Keep $R$ fixed and double the voltage. The current doubles.
• Keep $V$ fixed and double the resistance. The current becomes half.

To find the current, we write the law as $I = \dfrac{V}{R}$. Here is an example. A $12\ \text{V}$ battery is connected across a $4\ \Omega$ resistor. The current is $I = \dfrac{12\ \text{V}}{4\ \Omega} = 3\ \text{A}$.

Simulation: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0).

What decides the resistance of a wire?

A long wire is harder to push current through than a short one. A thin wire is harder than a thick one. So the resistance of a wire is

Here $L$ is the length, $A$ is the area of the cross section and $\rho$ (rho) is the resistivity. It depends on the material and on the temperature. It does not depend on the shape of the wire.

An example. A copper wire is $10\ \text{m}$ long and its cross section is $1\ \text{mm}^2$, which is $1 \times 10^{-6}\ \text{m}^2$. For copper, $\rho = 1.7 \times 10^{-8}\ \Omega\,\text{m}$.

What if we stretch the wire? Suppose a wire is pulled to twice its length. Its volume stays the same, so its area becomes half. Then $L$ becomes 2 times and $A$ becomes $\tfrac{1}{2}$. So $R$ becomes $2 \div \tfrac{1}{2} = 4$ times bigger.

Power in a resistor

A current in a resistor makes heat. The power is the energy per second.

A heater is connected to $230\ \text{V}$, and it draws $5\ \text{A}$. Its power is $P = VI = 230\ \text{V} \times 5\ \text{A} = 1150\ \text{W}$.

Resistors in series

Resistors in series are one after another. There is only one path for the current.

• The same current flows through every resistor.
• The voltage of the battery is shared: $V = V_1 + V_2 + V_3$.
• The resistances add: $R_{\text{eq}} = R_1 + R_2 + R_3$.

Adding a resistor in series always makes the total resistance bigger.

Resistors in parallel

Resistors in parallel are side by side. All of them are connected across the same two points.

• The same voltage is across every resistor.
• The current splits between the branches: $I = I_1 + I_2 + I_3$.
• The reciprocals add: $\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$.

For two resistors only, we can use $R_{\text{eq}} = \dfrac{R_1 R_2}{R_1 + R_2}$. Adding a resistor in parallel always makes the total resistance smaller. The current has one more path.

Simulation: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0).

Solving a circuit with series and parallel parts

We reduce the circuit step by step. Here is the plan.

1. Find the equivalent resistance of each parallel part, and each series part.
2. Repeat until there is one resistor.
3. Find the total current with $I = \dfrac{V}{R_{\text{eq}}}$.
4. Go back out. Find the voltage and the current in every part.

A circuit that cannot be reduced in this way needs Kirchhoff's rules. We meet them in the next lesson.

Practise with new numbers

Try it in Python

This program finds the current when a $12\ \text{V}$ battery is connected to one resistor at a time.

Now we join two resistors, $6\ \Omega$ and $3\ \Omega$. This program finds the equivalent resistance in series and in parallel.

Remember these

• $V = IR$. The same law gives $I = V/R$ and $R = V/I$.
• $R = \rho L / A$. Stretching a wire to twice the length makes its resistance 4 times bigger.
• $P = VI = I^2R = V^2/R$.
• Series: same current, resistances add. Parallel: same voltage, reciprocals add.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A heater has a resistance of $50\ \Omega$. It is joined to $230\ \text{V}$. What current does it draw, and what power does it use?

2. Resistors of $2\ \Omega$, $3\ \Omega$ and $6\ \Omega$ are joined in parallel. What is the total resistance?

3. A wire of $10\ \Omega$ is cut into two equal pieces, and the pieces are joined side by side, in parallel. What is the resistance now?

4. A $12\ \Omega$ and an $18\ \Omega$ resistor are in series across $15\ \text{V}$. What is the voltage across the $18\ \Omega$ one?