In english place

Here is the solution for question 3:

a) The property of the ammeter that enables it to be connected as shown is its very low internal resistance. This ensures that it does not significantly alter the total resistance or current in the circuit when connected in series.

b) Two advantages of connecting bulbs in parallel in our homes are:

• If one bulb blows or fails, the other bulbs in the circuit will continue to operate.
• Each bulb receives the full supply voltage, allowing them to operate at their intended brightness.

c)

(i) To calculate the combined resistance of the circuit:
The three $4 \, \Omega$ bulbs are connected in parallel. The formula for total resistance ($R_{total}$) in a parallel circuit is:

$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

Substitute the resistance of each bulb ($R_1 = R_2 = R_3 = 4 \, \Omega$):

$$\frac{1}{R_{total}} = \frac{1}{4 \, \Omega} + \frac{1}{4 \, \Omega} + \frac{1}{4 \, \Omega}$$ $$\frac{1}{R_{total}} = \frac{3}{4 \, \Omega}$$ $$R_{total} = \frac{4}{3} \, \Omega$$ $$R_{total} \approx \boxed{1.33 \, \Omega}$$

(ii) To calculate the current flowing through the ammeter:
The ammeter measures the total current ($I_{total}$) in the circuit. Using Ohm's Law, $V = I_{total} R_{total}$:

$$I_{total} = \frac{V}{R_{total}}$$

Substitute the supply voltage ($V = 240 \, \text{V}$) and the total resistance ($R_{total} = \frac{4}{3} \, \Omega$):

$$I_{total} = \frac{240 \, \text{V}}{\frac{4}{3} \, \Omega}$$ $$I_{total} = 240 \, \text{V} \times \frac{3}{4 \, \Omega}$$ $$I_{total} = \boxed{180 \, \text{A}}$$

(iii) To calculate the power consumed when all the bulbs are in use:
The total power consumed ($P_{total}$) can be calculated using the formula $P_{total} = V I_{total}$:

$$P_{total} = 240 \, \text{V} \times 180 \, \text{A}$$ $$P_{total} = \boxed{43200 \, \text{W}}$$