Question

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

Answer 1

Question:

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

Solution:

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

Answer 2

Given :-

• Resistance of electric lamp,$\rm\:R_1=100\Omega$

• Resistance of toaster,$\rm\:R_2=50\Omega$

• Resistance of water filter,$\rm\:R_3=500\Omega$

To find :-

• Resistance (R) = ?

Solution :-

we know that,

✭ ${\underline{\boxed{\bf\red{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}}}}$

$\rm\longrightarrow\:\dfrac{1}{R}=\dfrac{1}{100}+\dfrac{1}{50}+\dfrac{1}{500}$

$\rm\longrightarrow\:\dfrac{1}{R}=\dfrac{5+10+1}{500}$

$\rm\longrightarrow\:\dfrac{1}{R}=\dfrac{16}{500}$

$\rm\longrightarrow\:R=\dfrac{500}{16}$

$\rm\longrightarrow\:R=31.25\Omega$

Hence,the required resistance will be 31.25 Ω.

Now,

Given :-

• Potential difference (V) = 220 v

• Resistance (R) = 31.25 Ω (calculated in the above Solution)

To find :-

• Current (I) = ?

Solution :-

By Ohm's law,

we know that,

✭ ${\underline{\boxed{\bf\green{\dfrac{V}{I}=R}}}}$

$\rm\longrightarrow\:\dfrac{220}{I}=31.25$

$\rm\longrighrarrow\:I=\dfrac{220}{31.25}$

$\rm\longrightarrow\:I=7.04\:A$

Hence,the required current will be 7.04 amperes.