Question

An electric heater consists of three similar heating elements A, B and C, connected as shown in the figure above. 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.

1. Calculate the resistance of one heating element.
2. Calculate the current in each resistor when only S₁ and S₃ are closed.
3. Calculate the power dissipated across A when S₁, S₂ and S₃ are closed.

Answer 1

Given: Each heating element A, B and C is rated.

Power (P) = 1.2 kW = 1200 W

Voltage (V) = 240 V

Supply voltage = 240 V

i. We use formula,

P = `V^2/R`

⇒ R = `240^2/1200`

= `57600/1200`

= 48 Ω

∴ Resistance of one heating element is 48 Ω.

ii. For S₁ and S₃ are closed.

Current in C (connected directly across 240 V) using Ohm’s law.

V = IR

⇒ I = `V/R`

= `240/48`

= 5 A

∴ Current in C = 5 A

Current in A and B (connected in series):

Total resistance in series `(R_"series") = R_A + R_B`

= 48+ 48

= 96 Ω

I = `V/R`

= `240/96`

= 2.5 A

∴ Current through A and B = 2.5 A

iii. Power across A for S₁, S₂ and S₃ are closed: Now all three elements are connected directly to 240 V.

Current through A = 5 A (same as element C) using formula

P = I²R 

= (5)² × 48

= 25 × 48

= 1200

= 1.2 kW

∴ Power dissipated across A is 1.2 kW.