Science, asked by Anonymous, 22 days ago

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 ?​

Answers

Answered by Anonymous
1

Given parameters:

Resistance of electric lamp (R1) = 100 Ω

Resistance of toaster (R2) = 50 Ω

Resistance of water filter (R3) = 500 Ω

A 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.

Resistance 100 ohm, 50 ohm , 500 ohm connected in parallel

Solution

(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.

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Answered by LaRouge
1

Answer:

Given parameters:

Resistance of electric lamp (R1) = 100 Ω

Resistance of toaster (R2) = 50 Ω

Resistance of water filter (R3) = 500 Ω

A 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.

Resistance 100 ohm, 50 ohm , 500 ohm connected in parallel

Solution

(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.

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