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 bi as all three appliances?
ii) Write any two advantages of connecting electrical devices in parallel with the battery?
Answers
ii)There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage. The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
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.
The three resistors are connected in parallel as shown in the below figure.
(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:
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