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A current of 2 A is passed through a coil of resistance 75 ohm for 2 minutes. (a) How much heat energy is produced ? (b) How much charge is passed through the resistance?

Question

A current of 2 A is passed through a coil of resistance 75 ohm for 2 minutes. (a) How much heat energy is produced ? (b) How much charge is passed through the resistance?

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Solution 1

(a) The heat energy produced in a resistor can be calculated using Joule's law of heating, which states that the heat produced in a resistor is directly proportional to the square of the current passing through it, the resistance it offers to the current, and the time for which the current is passed. The formula is given by:

H = I^2 * R * t

where: H is the heat energy, I is the current, R is the resistance, and t is the time.

Substituting the given values:

H = (2 A)^2 * 75 ohm * (2 minutes * 60 seconds/minute) = 36000 Joules

(b) The amount of charge passed through the resistance can be calculated using the formula:

Q = I * t

where: Q is the charge, I is the current, and t is the time.

Substituting the given values:

Q = 2 A * (2 minutes * 60 seconds/minute) = 240 Coulombs

So, (a) the heat energy produced is 36000 Joules and (b) the charge passed through the resistance is 240 Coulombs.

This problem has been solved

Solution 2

(a) The heat energy produced in a resistor can be calculated using Joule's law of heating, which states that the heat produced in a resistor is directly proportional to the square of the current passing through it, the resistance of the circuit, and the time for which the current is passed. The formula is:

H = I^2 * R * t

where: H is the heat energy, I is the current, R is the resistance, and t is the time.

Substituting the given values:

H = (2 A)^2 * 75 ohm * 2 * 60 s = 36000 J

So, the heat energy produced is 36000 Joules.

(b) The charge passed through a resistor can be calculated using the formula:

Q = I * t

where: Q is the charge, I is the current, and t is the time.

Substituting the given values:

Q = 2 A * 2 * 60 s = 240 C

So, the charge passed through the resistance is 240 Coulombs.

This problem has been solved

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