An electric lamp of 100 ohm, a toaster of resistance 50 ohm, and a water filter
of resistance 500 ohm 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?
Answers
Given parameters:
Resistance of electric lamp (R1) = 100 Ω
Resistance of toaster (R2) = 50 Ω
Resistance of water filter (R3) = 500 Ω
Potential difference of the source (V) = 220 V
To find
(1) Resistance of an electric iron
(2) The flow of electric current through the electric iron.
(1) Total resistance (R) can be calculated as shown below.
1/R = 1/R₁ + 1/R₂ + 1/R₃
1/R = 1/100 + 1/50 + 1/500
1/R = ( 5 + 10 + 1 )/500
1/R = 16/500
500/16 = R
Total Resistance (R) = 500/16 Ω
R = 31.25 Ω
(2) According to ohm’s law the current through a conductor between two points is directly proportional to the voltage across the two points.
V ∝ I or V = IR
R is constant called resistance.
R = V/I
500/16 = 220/I
I = 220 × (16/500)
I = 7.04 A
Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.
Answer:
Resistance of electric lamp (R1) = 100 Ω
Resistance of toaster (R2) = 5 0 Ω
Resistance of water filter (R3) = 500 Ω
Potential difference of the source, V = 220 V
Since all the resistance are in parallel then the equivalent resistance R of the circuit will be
According to Ohm’s law
Ohm’s Law
Where current flowing through the circuit = I
All the three given appliances are drawing 7.04 A of current.
Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A
Let Rʹ be the resistance of the electric iron. According to Ohm’s law,
Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.