Question

an electric lamp of 100 ohm ,a toaster of resistance 50 ohm , and a water filter of resistance 500 ohm are connected in parraallel to a 220 volt sourc. what is the resistance of an elactric iron connected to the same source that takes as much current as all three appliances , and what is the current through it ?

































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Answer 1

Let R₁=100 Ω, R₂=50 Ω, R₃=500 Ω
V(Voltage of the circuit)=220 V
R(resistance of the iron)= ? (to be calculated)

By using the formula of parallel combination of resistors,
1/R=1/R₁ + 1/R₂ + 1/R₃
1/R=1/100 + 1/50 + 1/500
1/R=(5+10+1)/500  ⇒1/R=16/500  ⇒16R=500
                                                        ⇒R=500/16=31.25 Ω

I(Current flowing through the iron)=? (to be calculated)
By using the formula,
I=V/R
⇒I=220/31.25 =7.04 A

∴Resistance of the electric iron= 31.25 Ω
  Current flowing through the iron= 7.04 A

Answer 2

$\bf{\underline{\underline \green{AnswEr:-}}}$

In a parallel arrangement, the potential difference remain same but electric current is different.

Here, we have R₁ = 100 Ω, R₂ = 50 Ω and R₃ = 500 Ω

Total 1/R = 1/R₁ + 1/R₂ + 1/R₃

Total 1/R = 1/100 Ω + 1/50 Ω + 1/500 Ω

Total 1/R = ( 5 + 10 + 1 )/500

Total 1/R = 16/500

500/16 = Total R

Total Resistance = 500/16 Ω

Since, R = V/i

500/16 = 220/i

i = 220 × 16/500

i = 7.04 A

Here resistance and electric current used will be same.and therefore answers are :-

• [1] 500/16 Ω OR In Decimal, 31.25 Ω
• [2] 7.04 A