Ohm's law helps us find the unknown quantities in a circuit.

Ohm's law answers these questions:

Given a circuit with a particular voltage supplied by a battery or generator and a given resistance, what is the current?
Given a segment of a circuit with a given potential difference and given resistance, what is the current?
Given a resistor and a current, what is the potential difference across the resistor?

Here's how Ohm's law works:

Using I for current in amps, V for potential difference in volts, and R for resistance in ohms, Ohm's law is

I = V/R

It makes sense that the higher the voltage, the more current is pushed through a circuit, and the higher the resistance, the less current flows.

Example 1:
Finding current in a simple series circuit

$I = \frac{V}{R} or I = \frac{10 V}{50 Ω} = 0.2 A$

Example 2:
Finding current in each path of a parallel circuit.

The current through resistor 1 is given by Ohm's law:
$I_{1} = \frac{5 V}{30 Ω} = 0.17 A$

The current through resistor 2 is also given by Ohm's law:
$I_{2} = \frac{5 V}{60 Ω} = 0.083 A$

Note that the potential difference provided by the battery exists simultaneously across both parallel loops. The voltage drop of 5 V across the battery terminals exists across points A-D (thus across resistor 1) and also across points B-C. We assume in this problem that the bulb has no resistance, so all the voltage drop between B-C is across resistor 2.

The total current in the circuit is the sum of the current in each parallel loop. That is, the total current is the current through resistor 1 and the current through resistor 2, or $0.17 A + 0.083 A or 0.253 A$ .

Notice that more current flows through the path with the lower resistor. If you wanted to select the parallel loop with the highest current, you would choose the one with the lowest resistance.

Ohm's law shows how circuits can be designed to function on a practical level.

Generally, you will have a fixed voltage source (e.g., a 110 V outlet or a 9 V battery). Appliances function best with a certain level of current. Too much current can cause them to overheat and the wires to melt, and too little current may not be enough to drive the appliance or light the bulb. But given a voltage source, resistors can be used to set the current to the desired level.

Example 3:
Using resistance to set the current.

The current through a bulb should be no more than 0.5 A. Given a 10 V battery, what resistor should be inserted?

$I = \frac{V}{R} or R = \frac{V}{I} = \frac{10 V}{0.5 A} = 20 Ω$

Therefore a resistance of 20 ohms should be added to the circuit.

Example 4:
Inferring the voltage drop across a resistor

In a circuit path with a resistor of 35 $Ω$ , the current is measured with an ammeter and found to be 10 mA. What must be the potential difference or voltage drop across the resistor, to drive that current through that resistance?

$V = I R = 1 m A x 35 Ω = 1 x 10^{- 3} A x 35 V = 0.035 V$

Example 5:
A circuit with capacitors

A capacitor in "direct current" or DC circuits, such as we have studied, acts like a break in the circuit. (The situation is more complicated in an AC or alternating current circuit.)

In the circuit shown on top in the figure, the capacitor CI acts as a break in the circuit below resistor R2. No current can flow through the path with C1, so no current flows through R2. The circuit shown on the bottom is electrically equivalent to the circuit on the top.