Question

Two tungsten bulbs of power 50 W and 60 W work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?​

Answer 1

Answer:

Formula of power:

$P = \dfrac{V^2}{R}$

Calculating the resistance of first bulb we get:

$\implies 50 = \dfrac{220 \times 220}{R_1}\\\\\\\implies R_1 = \dfrac{48400}{50}\\\\\\\implies \boxed{ \bf{R_1 = 968 \:\:\Omega}}$

Hence resistance of Bulb 1 is 968 ohms.

Therefore for the second bulb we get:

$\implies 60 = \dfrac{ 220 \times 220}{R_2}\\\\\\\implies R_2 = \dfrac{48400}{60}\\\\\\\implies \boxed{ \bf{R_2 = 806.67 \:\:\Omega}}$

Hence the net resistance offered by the bulbs in the parallel circuit is:

$\implies \dfrac{1}{R_{net}} = \dfrac{1}{968} + \dfrac{1}{806.67}\\\\\\\implies \dfrac{1}{R_{net}} = \dfrac{ 806.67 + 968}{(968 \times 806.67)}\\\\\\\implies \dfrac{1}{R_{net}} = \dfrac{1774.67}{780856.56}\\\\\\\implies R_{net} = \dfrac{780856.56}{1774.67}\\\\\\\implies \boxed{ \bf{R_{net} = 440.000092 \approx 440 \:\:\Omega}}$

Hence the current flowing in the main conductor is calculated as:

$\implies I =\dfrac{V}{R}\\\\\\\implies I = \dfrac{220}{440}\\\\\\\implies I = \dfrac{1}{2}\\\\\\\implies \boxed{ \bf{ I = 0.5 \:A}}$

Answer 2

Answer:

$0.5A$

Explanation:

$P1 = 50W$

$R_{1} = \frac{ { \: \: v}^{2} }{ \: p _{1}} = \frac{ { \: \: (220)}^{2} }{50}$

$\frac{4840}{5} Ω$

$P _{2} = 60W$

$R _{2} = \frac{ \: {V}^{2} }{P _{2}} = \frac{ { \: (220)}^{2} }{60} \frac{ \: 4840 \: }{6} Ω$

$\frac {1}{R_{P} } = \frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac { \: 5 \: }{ \: 4840 \: } + \frac { \: 6 \: }{ \: 4840 \: }$

$R _{P} = 440 \: Ω$

$I = \frac{ \: V}{ \: R_{P}} = \frac{ \: 220 \: }{ \: 440 \: } A=0.5A$