A given conductor carrying a current of 1 A produces an amount of heat equal to 2000 J. If the current through the conductor is doubled, the amount of heat produced will be
8000 J
This problem involves calculating the heat produced in a conductor when the electric current flowing through it changes. The amount of heat produced by a conductor carrying current is described by Joule's Law of Heating.
Joule's Law states that the heat (\(H\)) produced in a resistor when a current (\(I\)) flows through it for a time (\(t\)) is directly proportional to the square of the current, the resistance (\(R\)) of the conductor, and the time for which the current flows. Mathematically, this is expressed as:
\(H = I^2 R t\)
In this problem, the conductor and the time duration are not mentioned to be changing. This means the resistance (\(R\)) of the conductor and the time (\(t\)) for which the current flows remain constant.
Therefore, the heat produced is directly proportional to the square of the current:
\(H \propto I^2\) (when R and t are constant)
We are given the initial conditions:
We are asked to find the heat produced (\(H_2\)) when the current is doubled.
Let's use Joule's Law for both cases:
For the initial case (current \(I_1\)):
\(H_1 = I_1^2 R t\)
We know \(H_1 = 2000\) J and \(I_1 = 1\) A. So,
\(2000 \text{ J} = (1 \text{ A})^2 R t\)
\(2000 = 1 \times R t\)
\(2000 = R t \quad \text{(Equation 1)}\)
For the new case (current \(I_2\)):
\(H_2 = I_2^2 R t\)
We know \(I_2 = 2 I_1 = 2 \times 1 = 2\) A. So,
\(H_2 = (2 \text{ A})^2 R t\)
\(H_2 = 4 R t \quad \text{(Equation 2)}\)
Now, we can substitute the value of \(R t\) from Equation 1 into Equation 2:
\(H_2 = 4 \times (R t)\)
\(H_2 = 4 \times 2000 \text{ J}\)
\(H_2 = 8000 \text{ J}\)
Alternatively, since \(H \propto I^2\) when \(R\) and \(t\) are constant, we can write the ratio:
\(\frac{H_2}{H_1} = \frac{I_2^2 R t}{I_1^2 R t}\)
\(\frac{H_2}{H_1} = \frac{I_2^2}{I_1^2} = \left(\frac{I_2}{I_1}\right)^2\)
We know \(I_2 = 2 I_1\), so \(\frac{I_2}{I_1} = 2\). Substituting this into the ratio equation:
\(\frac{H_2}{H_1} = (2)^2 = 4\)
\(H_2 = 4 \times H_1\)
\(H_2 = 4 \times 2000 \text{ J}\)
\(H_2 = 8000 \text{ J}\)
When the current through the conductor is doubled, the heat produced becomes four times the original heat.
The amount of heat produced when the current through the conductor is doubled will be 8000 J.
| Parameter | Initial Value | New Value |
|---|---|---|
| Current (\(I\)) | \(I_1 = 1\) A | \(I_2 = 2 I_1 = 2\) A |
| Heat Produced (\(H\)) | \(H_1 = 2000\) J | \(H_2 = ?\) |
| Resistance (\(R\)) | Constant | Constant |
| Time (\(t\)) | Constant | Constant |
| Concept | Description | Formula |
|---|---|---|
| Joule's Law of Heating | Relates heat produced to current, resistance, and time. | \(H = I^2 R t\) |
| Relation \(H \propto I^2\) | If R and t are constant, heat is proportional to the square of the current. Doubling current increases heat by \(2^2 = 4\) times. Tripling current increases heat by \(3^2 = 9\) times, and so on. | \(H_2 / H_1 = (I_2 / I_1)^2\) |
The heat produced in a conductor depends on several factors:
Understanding these factors is crucial for designing electrical circuits and appliances, especially concerning heat dissipation and safety.
Which one of the following is not a form of stored energy?
Which of the following is NOT a true difference between EMF and potential difference (PD)?
A battery of EMF 6.0 V and internal resistance 1.0 Ω is connected to a resistor of 11 Ω. The terminal potential difference for the battery is:
Consider two cells of emf ε1 and ε2 with internal resistances r1 and r2, respectively. The two cells are connected in parallel by connecting their positive terminals together and connecting their negative terminals together. The combination is equivalent to a single cell with emf given by:
Two batteries, E1 (emf: 3 V, internal resistance: 0.5 Ω) and E2(emf: 6 V, internal resistance: 1.0 Ω), are connected in series by connecting the positive terminal of E2 to the negative terminal of E1 . A third battery E3 (emf: 6 V, internal resistance: 1.0 Ω) is connected in parallel with this combination by connecting its positive terminal to the positive terminal of E1 and its negative terminal to the E2. The equivalent emf of this combination is: