Question

Two bubls rated 100W is at 220V and 200w at 220V are connected in parallel to a220V line. What total current is drawn by them?​

Answer 1

Explanation:

In parallel combination, voltage is constant.

Hence we can use the formula:

$\large \boxed{\boxed{\sf \bf P =\dfrac{V^2}{R}}}$

Let $\sf R_1$ and $\sf R_2$ be the resistances of the bulbs.

$\sf \bf :\longmapsto P_1 = \dfrac{V^2}{R_1}$

$\sf \bf :\longmapsto 100 = \dfrac{(220)^2}{R_1}$

$\sf \bf :\longmapsto R_1 = \dfrac{220 \times 220}{100}$

$\boxed{\sf{\bf{R_1 = 484}}\Omega}$

$\sf \bf :\longmapsto P_2 = \dfrac{V^2}{R_2}$

$\sf \bf :\longmapsto 200 = \dfrac{(220)^2}{R_2}$

$\sf \bf :\longmapsto R_2 = \dfrac{(220 \times 220)}{200}$

$\sf \bf :\longmapsto R_2 = 242 \Omega$

Now in parallel combination, the current is different for different resistances but coltage is same.

Ohm's law :

$\large{\boxed{\sf \bf V =IR}}$

Current in first hulb is given as:

$\sf \bf :\longmapsto I_1 = \dfrac{V}{R_1}$

$\sf \bf :\longmapsto I_1 =\dfrac{220}{484}$

$\boxed{\boxed{\sf \bf I_1 = \dfrac{5}{11} A}}$

Current in second bulb is given as:

$\sf \bf :\longmapsto I_2 = \dfrac{V}{R_2}$

$\sf \bf :\longmapsto I_2 = \dfrac{220}{242}$

$\boxed{\boxed{\sf \bf I_3 = \dfrac{10}{11} A}}$

Hence, total current drawn is,

$\sf \bf :\longmapsto I_1 + I_2$

$\sf \bf :\longmapsto \dfrac{5}{11} A + \dfrac{10}{11} A$

$\boxed{\boxed{\sf \bf :\longmapsto \dfrac{15}{11}A \sim 1.36 A}}$