Question

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Answer 1

Answer:

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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

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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Answer 2

Answer:

Given parameter:

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

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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