An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω 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

Resistance of electric lamp, *R _{1}* = 100 Ω

Resistance of toaster,

*R*= 50 Ω

_{2}Resistance of water filter,

*R*= 500 Ω

_{3}Potential difference of the source,

*V*= 220

*V*

These are connected in parallel, as shown in the following figure

Let *R* be the equivalent resistance of the circuit.

The net resistance in parallel is given by

1/R=1/R_1+1/R_2+1/R_3

Here, R_{1} = 100 Ω, R_{2} = 50 Ω and R_{3} = 500 Ω

`1/R=1/100+1/50+1/500`

`=(5+10+1)/500=16/500`

`R=500/16=31.25 Ω`

Now, using Ohm’s law V = IR, we have

I =V/R

`=(220 V)/(31.25 Ω)`

= 7.04 A

Hence, the resistance of electric iron is 31.25 Ω and current through it is 7.04 A.