The electric circuit (below figure) in a clothes dryer contains two heaters X and Y in parallel. The below figure shows the circuit connected to a 230 V power supply. When both switches are closed, the current in X is 3.5 A.

Analyse the circuit given above and answer the following questions.

(a) Calculate the power developed in heater X.

(b) If the resistance of X is double that of Y calculate the current in heater Y.

Given,

Voltage supply ($\text V$) = 230 V Current flowing through X ($\text I_\text X$) = 3.5 A

(a) Power developed ($\text P$) in heater X is given by,

$\text P = \text V \text I_\text X \\[1em] = 3.5 \times 230 \\[1em] \Rightarrow \text P = 805 \text{ W}$

Hence, the power developed in heater X is 805 W.

(b) Given,

Resistance of the heater X ($\text R$\text X) = 2 x Resistance of the heater Y ($\text R\text Y$)

Let, current flowing through the heater Y be '$\text I_\text Y$'.

Since, heaters X and Y are connected in parallel combination then potential drop across them should be equal to 230 V i.e.,

$\text V$\text X = \text V\text Y = 230 V …………… (i)

From Ohm's law,

$\text V = \text {IR}$

On putting in equation (i),

$\Rightarrow \text I$\text X\text R\text X = \text I\text Y\text R\text Y \\[1em] \Rightarrow \text I\text X \times 2\text R\text Y = \text I\text Y\text R\text Y \\[1em] \Rightarrow \text I\text Y = \dfrac{2\text I\text X \text R\text Y}{\text R\text Y} \\[1em] \Rightarrow \text I\text Y = \dfrac{2\text I\text X}{1} \\[1em] \Rightarrow \text I\text Y = 2 \times 3.5 \\[1em] \Rightarrow \text I\text Y = 7.0\ \text A

Hence, the current in heater Y is 7.0 A.