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
Answer
- 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
_________________________
( 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
_________________________
( 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
_________________________
Potential difference (V) = 240 V
Current (I) = 0.5 A
Let 'R' be the resistance.
Let 'H' be heat generated.
(a) According to the definition, one ampere means one coulomb of charge flowing per second.
Hence, 0.5 A implies 0.5 C charge per second.
(b) From Ohm's law,
V = IR
R = V/I = 240/0.5 = 480 ohm
(c) H = I²Rt
H = 0.5×0.5×480×300
=> 0.25×144000
=> 36,000 J