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?

Given,

Potential difference across circuit = 220 V

Resistance of electric lamp = 100 Ω

Resistance of toaster = 50 Ω

Resistance of water filter = 500 Ω

Equivalent resistance across the three appliances connected in parallel

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{100}} + \dfrac{1}{\text{50}}+ \dfrac{1}{\text{500}} = \dfrac{5+10+1}{\text{500}} = \dfrac{16}{\text{500}}$

Hence, R_p = $\dfrac{500}{\text{16}} = \dfrac{125}{\text{4}} = 31.25 \space \Omega$

The resistance of the electric iron connected to the same source will be 31.25 Ω as same current flows through it as that across the three appliances (connected in parallel).

Current across electric iron = ?

Applying Ohm's law

V = IR

Substituting we get,

220 = I x 31.25

I = $\dfrac{220}{31.25}$ = 7.04 A

Therefore, current through the electric iron will be 7.04 A.