Question

two lamps,one rated 100w at 220v and other 60w at 220v are connected in parallel to electric mains supply. what current is drawn from the line if the supply voltage is 220v.​

Answer 1

Given :

For first lamp :-

• Power = 100W
• Potential difference = 220V

For second lamp :-

• Power = 60W
• Potential difference = 220V

Both lamps are connected in parallel.

To find :

Current drawn from the line if the supply voltage is 220V.

Solution :

firstly we have to find resistance of Two lamps,

In first lamp, Let the resistance be R₁

we know ,

${ \leadsto{ \bf{Power, P = \dfrac{V^2}{R_1}}}}$

${ \leadsto{ \bf{100 = \dfrac{(220) ^{2} }{R_1}}}}$

${ \leadsto{ \bf{R_1 = \dfrac{48400}{100}}}}$

${ \leadsto{ \bf{R_1 = 484 \: \Omega}}}$

Similarly,

In second lamp, Let the resistance be R₂

${ \leadsto{ \bf{Power, P = \dfrac{V^2}{R_2}}}}$

${ \leadsto{ \bf{60 = \dfrac{ {(220)}^{2} }{R_1}}}}$

${ \leadsto{ \bf{R_2 = \dfrac{48400}{60}}}}$

${ \leadsto{ \bf{R_2 = 806.7 \: \Omega}}}$

The two lamps of resistance 484Ω and 806Ω are connected in parallel.

$\bigstar$ The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

${ \leadsto{ \bf{\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}}}$

${ \leadsto{ \bf{\dfrac{1}{R} = \dfrac{1}{484} + \dfrac{1}{806.7}}}}$

${ \leadsto{ \bf{\dfrac{1}{R} = \dfrac{806.7 + 484}{484 \times 806.7} }}}$

${ \leadsto{ \bf{\dfrac{1}{R} = \dfrac{1290.7}{390442.8} }}}$

${ \leadsto{ \bf{R = \dfrac{390442.8}{1290.7} }}}$

${ \leadsto{ \bf{R =302.5 \: \Omega}}}$

hence, total resistance, R = 302.5 Ω

Current drawn from the line can now be calculated by using ohm's law .i.e.,

$\bigstar$ At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. .i.e.,

${ \leadsto{ \bf{V = R \times I}}}$

here, V denotes potential difference, R denotes resistance and I denotes current.

${ \leadsto{ \bf{220 = 302.5 \times I}}}$

${ \leadsto{ \bf{ I = \dfrac{220}{302.5} }}}$

${ \leadsto{ \bf{I = 0.73A }}}$

Thus, the current drawn from line is 0.73A.

Remember !

SI unit of current is ampere (A)

SI unit of resistance is ohms (Ω)

SI unit of Power is watts (W)