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Lesson: Chapter 19 — Current Electricity

19.4. Ohm's law 5 of 5

Using V = IR

Rearranged, the definition of resistance becomes the formula you will use for the rest of the chapter.

The working form of Ohm's law

V = I R

Potential difference in volts, current in amperes, resistance in ohms.

Example 1

The problem

A bulb of resistance 6 Ω is connected into a circuit and a current of 1.5 A flows through it. Find the potential difference between its two ends.

Applying V = IR to the current flowing through the bulb:

V = 1.5 × 6

potential difference across the bulb = 9.0 V

Exercise 19.2

The problem

  1. A bulb is connected to a 12 V electricity supply and a current of 0.5 A flows through it. What is the resistance of the bulb's filament at that moment?
  2. A nichrome wire coil has a resistance of 10 Ω. When it is connected to an electricity supply a current of 0.6 A flows. What is the potential difference between the terminals of the supply the coil is connected to?
  3. A nichrome wire coil has a resistance of 6 Ω. When it is connected to a 3 V electricity supply, what current flows through it?

Which way to rearrange it

All three parts use the same formula and differ only in which letter is missing. Given V and I, divide to get R. Given I and R, multiply to get V. Given V and R, divide to get I.

One formula, three arrangements

V = IR is the only equation in this section. Everything else is knowing which of the three quantities the question has left out.