A 100W lamp bulb and a 4 kW water heater are connected to a 250V supply. Calculate
(a) the current which flows in each appliance
(b) the resistance of each appliance when
in use.
Ans.[(a)0.4 A, 16 A, (b) 625 Ω, 15.26 Ω]
Data Given:
Power of the Lamp Bulb = P₁ = 100 W
Power of the Heater = P₂ = 4 KW = 4 x 10³ W
Voltage of the power supply= V = 250 volts
Power of the Lamp Bulb = P₁ = 100 W
Power of the Heater = P₂ = 4 KW = 4 x 10³ W
Voltage of the power supply= V = 250 volts
To Find:
(a)
Current flowing through Lamp bulb = I₁ = ?
Current flowing through Heater = I₂ = ?
(b)
Resistance of Lamp bulb = R₁ = ?
Resistance of Heater = R₂ = ?
Solution:
(a)To find Current I₁ flowing through the Lamp bulb
by using formula P₁ = V I₁
I₁ = P₁V
by putting values
I₁ = 100W250V
I₁ = 0.4 A---------------Ans. 1(a)
To find Current I₂ flowing through the heater
by using formula P₂ = V I₂
I₂ = P₂V
by putting values
I₂ = 4x10³W250V
I₂ = 0.016 x 10³ W
I₂ = 16 x 10⁻³x 10³ W
I₂ = 16 A -----------------Ans.2(a)
by using formula P₁ = V I₁
by using formula P₂ = V I₂
I₂ = 16 x 10⁻³x 10³ W
(b)According to Ohm's Law
V = I R
or
R = VI
To find the resistance of Lamp bulb = R₁
R₁ = VI₁
R₁ = 250V0.4A
R₁ = 625 Ω ----------------Ans. 1(b)
To find the resistance of heater = R₂
R₂ = VI₂
R₂ = 250V16A
R₂ = 15.625 Ω -----------------Ans. 2(b)
(b)
According to Ohm's Law
V = I R
or
R = VI
To find the resistance of Lamp bulb = R₁
R₁ = VI₁
R₁ = 250V0.4A
R₁ = 625 Ω ----------------Ans. 1(b)
To find the resistance of heater = R₂
R₂ = VI₂
R₂ = 250V16A
R₂ = 15.625 Ω -----------------Ans. 2(b)
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