Question

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?​

Answer 1

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

________________________________

HOPE IT HELPS!:)

Answer 2

Answer:

Deleted account

2 days ago

Math

Secondary School

+50 pts

Answered

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?

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Answers

smitaprangya98

Smitaprangya98Ace

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

________________________________

( 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

Step-by-step explanation:

hope this helps please follow