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

Explanation

Resistance of electric lamp, `R_1=100Omega`
Resistance of toaster `R_2=50Omega`
Resistance of water filter, `R_3=500Omega`
Potential difference of the source `V=200V` these are connected in parallel, as shown in the following figure

Let R be the equivalent resistance of the circuit
Potential difference=220V
Sum of resistance can be calculated as follows
`1/R=1/(100Omega)+1/(50Omega)+1/(500Omega)`
`=(5+10+1)/(500)Omega=(16)/(500)Omega`
Therefore R`=(500)/(16)Omega=31.25Omega`
Electric current (1) `=V/R`
`implies1=(220V)/(31.25Omega)=7.044`
Since the electric iron takes same amount of current as three appliances, thus electric current through it =7.04A
The electric current=7.04A and potential difference=220V
Thus, Resistance of electric iron=Total resistance of three appliances=`31.25Omega`
Thus, electric current through the electric iron=7.04A
Resistance of electric iron=`31.25Omega`