2 The voltage supplied to a circuit is 240 V and the resistance is 60 Ω. Calculate the current flowing in the circuit.

Ohm's Law states that $V = IR$, where $V$ is voltage, $I$ is current, and $R$ is resistance. To find the current, we rearrange the formula to:

$$I = \frac{V}{R}$$

Given $V = 240$ V and $R = 60 \, \Omega$.

$$I = \frac{240\, \text{V}}{60\, \Omega}$$

$$I = 4\, \text{A}$$

The current flowing in the circuit is $\boxed{\text{4 A}}$.

3.3
A kettle operates at 12 A when connected to a 240 V supply. Calculate the resistance of the kettle.

Using Ohm's Law, $V = IR$, we can rearrange to find resistance:

$$R = \frac{V}{I}$$

Given $V = 240$ V and $I = 12$ A.

$$R = \frac{240\, \text{V}}{12\, \text{A}}$$

$$R = 20\, \Omega$$

The resistance of the kettle is $\boxed{\text{20 } \Omega}$.

QUESTION 4: REASONING AND INTERPRETATION

4.1 A resistor has a constant resistance of $10 \, \Omega$.

4.1.1 Calculate the current when the voltage is 20 V.

Given $V = 20$ V and $R = 10 \, \Omega$.

$$I = \frac{20\, \text{V}}{10\, \Omega}$$

$$I = 2\, \text{A}$$

The current is $\boxed{\text{2 A}}$.

4.1.2 Calculate the current when the voltage is increased to 40 V.

Given $V = 40$ V and $R = 10 \, \Omega$.

$$I = \frac{40\, \text{V}}{10\, \Omega}$$

$$I = 4\, \text{A}$$

The current is $\boxed{\text{4 A}}$.

4.1.3 What happens to the current when the voltage is doubled?

From 4.1.1, when $V = 20$ V, $I = 2$ A. From 4.1.2, when $V = 40$ V (which is double 20 V), $I = 4$ A (which is double 2 A).
When the voltage is doubled, the current also doubles. This is because current is directly proportional to voltage when resistance is constant.