Question

Two bulbs rate 200W at 220V and 400W at 220V are connected in parallel to a 220V line. What total current is drawn by them​

Answer 1

⇝ Given :-

• Power of 1st Bulb = P₁ = 200 W

• Power of 2nd Bulb = P₂ =400 W

• Potential Diff. = V = 220V

⇝ To Find :-

• The Total Current Drawn By the Bulbs .

⇝ Solution :-

We Know,

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

⭐ Using above formula we have to calculate Resistance of Both Bulbs.

❒ Calculating Resistance of 1st Bulb :

For First Bulb :

$200 = \dfrac{220 {}^{2} }{R_1}$

$:\longmapsto R_1= \dfrac{48400}{200}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf R_1 = 242 Ω} }}}$

❒ Calculating Resistance of 2nd Bulb :

For 2nd Bulb :

$400 = \dfrac{ {220}^{2} }{R_2}$

$:\longmapsto R_2 = \frac{48400}{400}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf R_2 =121 Ω} }}}$

❒ Calculating Net Resistance :

As Both bulbs are in parallel :

$\bf\dfrac{1}{R_{net}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$:\longmapsto\sf\dfrac{1}{R_{net}} = \frac{1}{242} + \frac{1}{221}$

$:\longmapsto \sf\dfrac{1}{R_{net}} = \frac{1 + 2}{242}$

$:\longmapsto\sf\dfrac{1}{R_{net}} = \frac{3}{242}$

$\pink{ \large :\longmapsto  \underline {\boxed{{\bf R_{net} = \frac{242}{3} Ω} }}}$

Now,

❒ Using Ohm's Law :

$:\longmapsto\bf V=IR$

$:\longmapsto220 = \text I \times \frac{242}{3}$

$:\longmapsto \text I = \dfrac{220 \times 3}{242}$

$\pink{ \large :\longmapsto  \underline {\boxed{\bf I= 2.72 \: A}}}$

Answer 2

• Power of first bulb=P1=200W
• Power of second bulb=P2=400W
• Potential difference of each bulb=220V

Let

• Resistances of two bulbs be R1 and R2 respectively

We know

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

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

$\\ \sf\longmapsto R_1=\dfrac{220^2}{200}$

$\\ \sf\longmapsto R_1=\dfrac{48400}{200}$

$\\ \bf\longmapsto R_1=242\Omega$

And

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

$\\ \sf\longmapsto R_2=\dfrac{220^2}{400}$

$\\ \sf\longmapsto R_2=\dfrac{48400}{400}$

$\\ \sf\longmapsto R_2=121\Omega$

Both are connected in parallel

$\boxed{\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}$

$\\ \sf\longmapsto \dfrac{1}{R_{eq}}=\dfrac{1}{242}+\dfrac{1}{121}$

$\\ \sf\longmapsto \dfrac{1}{R_{eq}}=\dfrac{1+2}{242}$

$\\ \sf\longmapsto \dfrac{1}{R_{eq}}=\dfrac{3}{242}$

$\\ \sf\longmapsto R_{eq}=\dfrac{242}{3}\Omega$

• Current=I

Using ohms law

$\boxed{\sf \dfrac{V}{I}=R}$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{220}{\dfrac{242}{3}}$

$\\ \sf\longmapsto I=\dfrac{220\times 3}{242}$

$\\ \sf\longmapsto I=\dfrac{660}{242}$

$\\ \sf\longmapsto I=2.72A$