An electric heater consists of three similar heating elements A, B and C, connected as shown in the figure below. Each heating element is rated as 1.2 kW, 240 V and has constant resistance. S₁, S₂ and S₃ are respective switches.

The circuit is connected to a 240 V supply.

(a) Calculate the resistance of one heating element.

(b) Calculate the current in each resistor when only S₁ and S₃ are closed.

(c) Calculate the power dissipated across A when S₁, S₂ and S₃ are closed.

Given,

• Power rating of each heater ($\text P$) = 1.2 kW = 1.2 x 1000 W = 1200 W
• Voltage rating of the heater ($\text V$) = Voltage supply = 240 V

(a) Resistance of one heating element is given by,

$\text R = \dfrac{\text V^2}{\text P} \\[1em] = \dfrac{240^2}{1200} \\[1em] = \dfrac{240\times 240}{1200} \\[1em] = \dfrac{240}{5} \\[1em] \Rightarrow \text R = 48\ \text Ω$

Hence, the resistance of one heating element is 48 Ω

(b) When S₁ and S₃ are closed then heating elements A and B are in series and this whole arrangement is in parallel combination with C.

Then,

Potential difference across A and B = Potential difference across C = 240 V

Now,

$\text {Current through C} = \dfrac{\text V}{\text R} \\[1em] = \dfrac{240}{48} \\[1em] = \dfrac{10}{2} \\[1em] = 5\ \text A$

Since A and B are connected in series then their effective resistance is given by,

$\text R_\text S = \text R + \text R \\[1em] = 48 + 48 \\[1em] = 96\ \text Ω$

Then,

$\text {Current through A and B} = \dfrac{\text V}{\text R_\text S} \\[1em] = \dfrac{240}{96} \\[1em] = \dfrac{10}{4} \\[1em] = 2.5\ \text A$

Hence, current through heating elements A nd B is 2.5 A and through C is 5 A.

(c) When S₁, S₂ and S₃ are closed then current only flows through A but no current flows through B since current takes path of minimum resistance.

So elements A and C are in parallel and have same resistance it means current flowing through them will be equal.

Current through A = Current through C = 5 A

Then,

$\text {Power dissipated across A} = \text {(Current through A)}^2 \times \text R \\[1em] = 5^2\times 48 \\[1em] = 25\times 48 \\[1em] = 1200\ \text W$

Hence, power dissipated across A is 1200 W.