Question

A heater is designed to operate with a power of $1000W$ on a line of $100V$. It is connected in combination with resistance of $10\Omega$ and a resistance R to line of $100V$. The value of R so the entire circuit operates with a power of $625W$ is

• A. $5\Omega$
• B. $10\Omega$
• C. $15\Omega$
• D. $20\Omega$

Answer 1

Answer: (C)

$15\Omega$

Power of heater, $P=1000W$ and potential difference is $V=100V$.
Thus, resistance of heater is: $R_1=\frac{V^2}{P}=\frac{{100}^2}{1000}=10\Omega$
Now it is connected in series to parallel combination of $R$ and $10\Omega$ resistor.
Thus the equivalent resistance of the whole circuit is $R_2=10+R||10=10+\frac{10R}{10+R}=\frac{100+20R}{10+R}$
Now the power, $\frac{V^2}{R_2}=P^{\prime }=625$
$R_2=\frac{{100}^2}{625}=16$

$\frac{100+20R}{10+R}=16$

$100+20R=160+16R$

$4R=60\Rightarrow R=15\Omega$