Question

A motor that pumps up water to a tank requires 10kW power. Its operating voltage is 220V. The only available resistor has a resistance of $11\Omega$. We know that we must use n such resistors in series and p such series resistor combinations in parallel. The values of n and p are respectively

• A. 25 and 16
• B. 21 and 11
• C. 11 and 25
• D. 16 and 25

Answer 1

The correct option is C 11 and 25

$P=\frac{V^2}{R}Hence,10\times {10}^3=\frac{{220}^2}{R}R=\frac{{220}^2}{10\times {10}^3}=\frac{48400}{10000}=4.84\Omega$
Hence the required resistance is $4.84\Omega$
Now, we know that we need an effective resistance of $4.84\Omega$
Given that $11\Omega$ resistors are arranged in the fashion shown below

When n = and p = 25,means that there are eleven resistors in series and 25 such series combinations in parallel.
$R_{eq}=R_1+R_2+R_3+........+R_n$is the formula for equivalent resistance for resistors in series
The equivalent resistance of the series combination will be 11 $\Omega +11\Omega +11\Omega .....11times=11\Omega \times 11=121\Omega$
$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+........+\frac{1}{R_n}$ will be the formula for equivalent resistance for resistors in parallel
The equivalent resistance will be the equivalent resistance of $25121\Omega$ resistances in parallel
Which is given by
$\frac{1}{R_{eq}}=\frac{1}{121}+{\frac{1}{121}}_2+\frac{1}{121}+.......25times$
$R_{eq}=\frac{121}{25}=4.84\Omega$, which is the required resistance.