Question

Several electric bulbs designed to be used on a 220V electric supply line are rated 10W.How many lamps can be connected in series with each other across the two wires of 220V if the maximum allowable current is 5A

Answer 1

I=5A

➡️V=220V

➡️P=VI

➡️P=5×220

➡️P=1100W

➡️Power of each bulb=10W

➡️No. of bulb=1100/10=110

➡️➡️Therefore numberof bulb is 110.

Make me brainliest:

Answer 2

FOR ONE BULB:-

Power, P = 10W

Potential Difference, V = 220V

Using the relation for R, we have

$\longrightarrow$$\large\tt\red{R=\frac{{V}^{2}}{P}}$

$\longrightarrow$$\large\tt\red{\frac{ {(220V)}^{2} }{10}}$

$\longrightarrow$$\large\tt\red{4840\mho}$

$\small\tt\green{Let\:the\:total\:number\:of\: bulbs\:be\:x}$

GIVEN THAT:-

$\large\sf{Current,I=5A}$

$\large\sf{Potential\:Difference,V=220V}$

$\huge\mathbb\orange{A/C,OHM'S\:LAW,}$

$\longrightarrow$$\large\tt\blue{R=\frac{V}{I}}$

$\longrightarrow$$\large\tt\blue{\frac{220}{5}}$

$\longrightarrow$$\large\tt\blue{44\mho}$

Now, for x times of bulbs of resistance $\small\sf{176\mho}$ , the equivalent resistance of the resistors connected in parallel is $\small\sf{44\mho}$

$\large\tt\pink{\frac{1}{44} = \frac{1}{4840} + \frac{1}{4840} + \frac{1}{4840} + ...to \: x \: times}$

$\large\tt\pink{\frac{1}{44}=\frac{x}{4840}}$

$\large\tt\pink{x=\frac{4840}{44}=110}$

$\therefore$ 110 bulbs of $\small\sf{4840\mho}$ are required to draw the given amount of current.