Physics

What are current, voltage and resistance?

Electric current (I) is the rate of flow of electric charge through a conductor. It is measured in ampere (A).

Potential difference or voltage (V) is the push that makes the charge flow. It is measured in volt (V).

Resistance (R) is the opposition that a conductor offers to the flow of current. It is measured in ohm (Ω).

Ohm's law

At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its two ends.

V = I × R
I = V / R
R = V / I

Here V is in volt, I is in ampere and R is in ohm. If you draw a graph of V (on the vertical axis) against I (on the horizontal axis) for a metallic conductor at constant temperature, you get a straight line through the origin. The slope of this line is the resistance.

Factors on which resistance depends

The resistance of a conductor depends on four things:

1. Length (l): resistance is directly proportional to the length.
2. Area of cross-section (A): resistance is inversely proportional to the area.
3. Material: every material has its own resistivity (ρ), also called specific resistance.
4. Temperature: for metals, resistance increases when the temperature increases.

These are combined in one formula:

R = ρ × l / A

The unit of resistivity is ohm-metre (Ω·m).

Resistors in series and in parallel

Series connection: the same current flows through every resistor, and the voltage is shared between them.
Total resistance R = R1 + R2 + R3

Parallel connection: the same voltage is across every resistor, and the current is shared between them.
1/R = 1/R1 + 1/R2 + 1/R3
The total resistance in parallel is always less than the smallest resistor.

Solved example 1: finding current

A potential difference of 12 V is applied across a resistor of 4 Ω. Find the current.

Given: V = 12 V, R = 4 Ω
To find: I
Solution: I = V / R = 12 / 4 = 3 A
Answer: The current is 3 A.

Solved example 2: resistors in series

Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series with a 20 V battery. Find (a) the total resistance, (b) the current and (c) the voltage across the 5 Ω resistor.

(a) R = 2 + 3 + 5 = 10 Ω
(b) I = V / R = 20 / 10 = 2 A
(c) V = I × R = 2 × 5 = 10 V

Check: the voltages across the three resistors are 4 V, 6 V and 10 V, and 4 + 6 + 10 = 20 V. This matches the battery voltage.

Solved example 3: resistors in parallel

Two resistors of 6 Ω and 3 Ω are connected in parallel across a 6 V battery. Find the total resistance and the total current drawn from the battery.

1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
So R = 2 Ω
I = V / R = 6 / 2 = 3 A

Check: the current through 6 Ω is 6/6 = 1 A and through 3 Ω is 6/3 = 2 A. Their sum is 3 A.

Solved example 4: change the units first

A small device draws a current of 500 mA from a 5 V supply. Find its resistance.

Given: I = 500 mA = 500 / 1000 = 0.5 A, V = 5 V
Solution: R = V / I = 5 / 0.5 = 10 Ω
Answer: The resistance is 10 Ω.

Many students lose marks here by using 500 instead of 0.5. Always convert mA to A before you put the values in the formula.

Solved example 5: using resistivity

A copper wire is 2 m long and has an area of cross-section of 1 × 10⁻⁶ m². The resistivity of copper is 1.7 × 10⁻⁸ Ω·m. Find the resistance of the wire.

R = ρ × l / A
R = (1.7 × 10⁻⁸ × 2) / (1 × 10⁻⁶)
R = 3.4 × 10⁻² Ω = 0.034 Ω

Remember: if the area is given in mm², convert it first. 1 mm² = 10⁻⁶ m².

Common mistakes to avoid

1. Writing the answer without a unit (A, V or Ω).
2. Not converting mA to A, or mm² to m².
3. Adding the resistances directly in a parallel connection.
4. Using the battery voltage for one resistor in a series circuit. Use V = I × R for that resistor only.
5. Drawing the V–I graph with the axes swapped.

Practice questions

1. A resistor of 10 Ω carries a current of 0.5 A. Find the potential difference across it.
2. A bulb draws 0.2 A from a 220 V supply. Find the resistance of the bulb.
3. Three resistors of 4 Ω, 6 Ω and 10 Ω are connected in series with a 10 V battery. Find the current.
4. Find the total resistance when 12 Ω and 4 Ω are connected in parallel.
5. A wire of resistance 5 Ω is cut into two equal halves. Find the resistance of each half, and then the resistance when the two halves are joined in parallel.

Answers:
1. V = I × R = 0.5 × 10 = 5 V
2. R = V / I = 220 / 0.2 = 1100 Ω
3. R = 20 Ω, so I = 10 / 20 = 0.5 A
4. 1/R = 1/12 + 3/12 = 4/12, so R = 3 Ω
5. Each half is 2.5 Ω. In parallel: (2.5 × 2.5) / (2.5 + 2.5) = 1.25 Ω

Quick revision

V = I × R
R = ρ × l / A
Series: R = R1 + R2 + R3
Parallel: 1/R = 1/R1 + 1/R2 + 1/R3

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