An electric lamp of resistance 100Ω, a toaster of resistance 50Ω and a water filter of resistance 500Ω 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: Resistance of electric lamp, R1 = 100Ω
Resistance of toaster, R2 = 50Ω
Resistance of water filter, R3 = 500Ω
Voltage, V = 220 V
Let Rp be the resistance of electric iron. Then:
\(\frac{1}{R_p}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)
[∵ Appliances are connected in parallel]
= \(\frac{1}{100}\) + \(\frac{1}{50}\) + \(\frac{1}{500}\)
= \(\frac{4}{125}\)
∴ Rp = \(\frac{125}{4}\) = 31.25Ω
Current through the electric iron,
I = \(\frac{V}{R_p}\) = \(\frac{220}{\frac{125}{4}}\)
= \(\frac{220 \times 4}{125}\) = 7.04 A