An electric lamp of `100 Omega`, a toaster of resistance `50 Omega` and a water filter of resistance `500 Omega` 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 ?

Options

A. R = 31.25 ohm , I = 7.04A
B. R = 35.25 ohm , I = 7.04A
C. R = 45.50 ohm , I = 5.09A
D. R = 25.25 ohm , I = 40.04A

Answer

A

Explanation

The combined resistance R of three resistors (or electrical devices) `R_(1),R_(2)` and `R_(3)`, connected in parallel is given by the formula:
`(1)/(R) = (1)/(R_(1)) +(1)/(R_(2))+(1)/(R_(3))`
Here, Resistance of electric lamp, `R_(1) = 100 Omega`
Resistance of toaster, `R_(2) = 50Omega`
And, Resistance of water filter, `R_(3) = 500 Omega`
Putting these values of `R_(1),R_(2)` and `R_(3)` in the above formula, we get:
`(1)/(R) = (1)/(100) +(1)/(50) +(1)/(500)`
`(1)/(R) = (5+10+1)/(500)`
`(1)/(R) = (16)/(500)`
`R = (500)/(16)`
`R = 31.25 Omega`
Thus, the resistance of electric iron will also be `31.25 Omega`.
Now, Potential difference, `V = 220 V`
Current, `I = ?` (To be calculated)
And, Resistance, `R = 31.25 Omega` (Calculated above)
By Ohm's law: `(V)/(I) = R`
`(220)/(I) = 31.25`
`I = (220)/(31.25)`
`I = 7.04 A`
Thus, the current passing through the electric iron is 7.04 amperes.