Question

A heater is operated with a power of $1000 \,W$ in a $100\, V$ line. It is connected in combination with a resistance of $10 \Omega$ and a resistance $R$ to a $100\, V$ line as shown in figure. What should be the value of $R$ so, that the heater operates with a power of $62.5$ watt.

• A. $10 \Omega$
• B. $62.5 \Omega$
• C. $1/5 \Omega$
• D. $5 \Omega$

Answer 1

Answer: (D)

Resistance of heater,
$R_{H}=\frac{V^{2}}{P}=\frac{\left(100\right)^{2}}{1000}=10\Omega$
Total resistance of circuit,
$R'=10+\left(\frac{10\times R}{10+R}\right)=\frac{100+20\,R}{10+R}$
current in heater,
$I_{H}=I\times\frac{R}{\left(10+R\right)}\times\frac{100}{\left(\frac{100+R}{10+R}\right)}\times\frac{R}{\left(10+R\right)}=\frac{5\times R}{5+R}$
So, Power,
$P=I^{2}_{H} R_{H}=\frac{25\,R}{\left(5+R\right)^{2}}\times10=62.5$
$\therefore R=5\,\Omega$