Question

class 10 question -36. (a) Two lamps, one is rated 100 W at 220 V, and the other 60 W at 220

V, are connected in parallel to a 220 V supply. Find the current drawn

from the supply line.

(b) Derive an expression when three resistors R1 , R2 and R3 are

connected in parallel in an electric circuit.​

Answer 1

$\huge\sf\pink{Answer}$

☞ Current in the circuit

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$\huge\sf\blue{Given}$

✭ Qn 1- Two lamps are rated 100 W & 60 W with a pd of 220 V

✭ Qn 2 - There are three resistors connected in parallel

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$\huge\sf\gray{To \:Find}$

◈ Qn 1 - Current in the circuit?

◈ Qn 2 - Derivation of the formula to find net resistance?

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$\huge\sf\purple{Steps}$

Question 1

We shall find the current through the circuit with the help of,

$\underline{\boxed{\sf P = VI}}$

• P = 100+60 = 160 W
• V = 220
• I = ?

Substituting the given values,

➝ $\sf P = VI$

➝ $\sf \dfrac{P}{V} = I$

➝ $\sf \dfrac{160}{220} = I$

➝ $\sf \red{I = 0.72 \ A}$

Question 2

Given that there are 3 Resistors - R¹,R² & R³ then,

◕ V = constant

◕ $\sf I = I_1 + I_2 + I_3 \:\:\: -eq(1)$

As per Ohm's law,

$\underline{\boxed{\sf V = IR}}$

It can also be written as,

➳ $\sf \dfrac{V}{R} = I$

⪼ $\sf I_1 = \dfrac{V}{R_1}$

⪼ $\sf I_2 = \dfrac{V}{R_2}$

⪼ $\sf I_3 = \dfrac{V}{R_3}$

Substituting the values in eq(1)

➳ $\sf I = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3}$

➳ $\sf I = V\bigg\lgroup\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\bigg\rgroup$

➳ $\sf \dfrac{I}{V} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$

➳ $\sf \orange{\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

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