Question

A heater coil operates at $1200v$ when connected to a $240W$ supply.The resistivity and area of across section of the coil are :

• A. $A=\frac{\rho l}{7000}$
• B. $A=\frac{\rho l}{8000}$
• C. $A=\frac{\rho l}{6000}$
• D. $A=\frac{\rho l}{600}$

Answer 1

Answer: (C)

$A=\frac{\rho l}{6000}$

We know Power$P=I\times V$-------------(1)_

Also $V=IR$, thus $I=\frac{V}{R}$ where I=current, V=potential difference and R=resistance.

From above relation equation 1 can be written as

$P=\frac{V^2}{R}$

$R=\frac{{1200}^2}{240}$

$R=6000\Omega$

As we know resistance is inversely proportional to the area of cross section

$R\alpha \frac{1}{A}$

$R=\rho \frac{l}{A}$

$A=\frac{\rho l}{R}$

$A=\frac{\rho l}{6000}$