A heater working in 240v mains consumers 0.5A current (a) calculate the quantity of charge flowing through the heater coil(b) calculate the resistance of the heater.(c) find the quantity of heat generated in the heater when worked for 5 minutes?
Answers
Step-by-step explanation:
Charge , Q = 0.5 C
Resistance , R = 480 Ω
Heat , H = 36 k-J
Given
A heater working in 240 V mains consumers 0.5 A current
To Find
( a ) calculate the quantity of charge flowing through the heater coil
( b ) calculate the resistance of the heater
( c ) find the quantity of heat generated in the heater when worked for 5 minutes
Concept Used
( a ) Ampere :
It is defined as current flows with electric charge of 1 Columb per second
1 A = 1 C/s
( b ) Ohm's Law
Electric current is directly proportional to voltage and inversely proportional to the resistance
I = V/R
V = IR
( c ) Joule's Law
It states the heat produced by an electric current , I flowing through he resistance , R for a time , t is equals to I²Rt
H = I²Rt
Solution
( a )
Current , I = 0.5 A = 0.5 C/s
So , the quantity of charge flowing through the heater coil is 0.5 C
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( b )
Voltage , V = 240 V
Current , I = 0.5 A
Resistance , R = ? Ω
Apply Ohm's Law ,
⇒ V = IR
⇒ 240 = 0.5 × R
⇒ R = 480 Ω
So , Resistance of the heater , R = 480 Ω
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( c )
Current , I = 0.5 A
Resistance , R = 480 Ω
Time , t = 5 min = 300 s
Quantity of heat , H = ? J
Apply Joules Law ,
⇒ H = I²Rt
⇒ H = (0.5)²×(480)×(300)
⇒ H = 36,000 J
⇒ H = 36 k-J
Step-by-step explanation:
Q=0.5
R=480
H=36
SOL-
CURRENT