Heat Dissipation in a Wire

This problem involves calculating the total heat dissipated in a wire when the charge flowing through it is time-dependent. To find the heat dissipated, we need to first determine the current flowing through the wire and then integrate the power dissipated over the time for which the current flows.

Current Calculation from Charge

The charge flowing through the resistor is given as a function of time \(t\):

\[ Q(t) = at - bt^2 \]

Where \(a\) and \(b\) are constants.

The electric current \(I\) is defined as the rate of flow of charge. Therefore, we can find the current by differentiating the charge function with respect to time:

\[ I(t) = \frac{dQ}{dt} \]

Differentiating the given charge equation:

\[ I(t) = \frac{d}{dt}(at - bt^2) \] \[ I(t) = a - 2bt \]

Time Interval for Current Flow

For heat to be dissipated, there must be a current flowing through the wire. The current \(I(t)\) starts at \(t=0\) and decreases with time. The current stops flowing when \(I(t)\) becomes zero.

Set \(I(t) = 0\) to find the time \(t_{max}\) when the current ceases:

\[ 0 = a - 2bt_{max} \] \[ 2bt_{max} = a \] \[ t_{max} = \frac{a}{2b} \]

So, the current flows from \(t = 0\) to \(t = \frac{a}{2b}\). We will integrate over this time interval.

Heat Dissipation Calculation

The heat \(H\) dissipated in a resistor \(R\) over a time interval is given by Joule's Law of Heating, which is the integral of instantaneous power dissipated. The instantaneous power \(P\) in a resistor is given by \(P = I^2R\).

\[ H = \int P \, dt = \int I^2 R \, dt \]

Substitute the expression for current \(I(t)\) and integrate from \(t=0\) to \(t=\frac{a}{2b}\):

\[ H = \int_{0}^{a/2b} (a - 2bt)^2 R \, dt \]

Since \(R\) is a constant, we can take it out of the integral:

\[ H = R \int_{0}^{a/2b} (a - 2bt)^2 \, dt \]

Expand the term \((a - 2bt)^2\):

\[ (a - 2bt)^2 = a^2 - 2(a)(2bt) + (2bt)^2 = a^2 - 4abt + 4b^2t^2 \]

Now, substitute this back into the integral:

\[ H = R \int_{0}^{a/2b} (a^2 - 4abt + 4b^2t^2) \, dt \]

Perform the integration term by term:

\[ H = R \left[ a^2t - 4ab\frac{t^2}{2} + 4b^2\frac{t^3}{3} \right]_{0}^{a/2b} \] \[ H = R \left[ a^2t - 2abt^2 + \frac{4}{3}b^2t^3 \right]_{0}^{a/2b} \]

Now, evaluate the expression at the upper limit \(t = \frac{a}{2b}\) and subtract its value at the lower limit \(t = 0\). The value at \(t=0\) will be zero for all terms.

Substitute \(t = \frac{a}{2b}\):

• First term: \( a^2\left(\frac{a}{2b}\right) = \frac{a^3}{2b} \)
• Second term: \( -2ab\left(\frac{a}{2b}\right)^2 = -2ab\left(\frac{a^2}{4b^2}\right) = -\frac{2a^3b}{4b^2} = -\frac{a^3}{2b} \)
• Third term: \( \frac{4}{3}b^2\left(\frac{a}{2b}\right)^3 = \frac{4}{3}b^2\left(\frac{a^3}{8b^3}\right) = \frac{4a^3b^2}{24b^3} = \frac{a^3}{6b} \)

Combine these terms:

\[ H = R \left[ \frac{a^3}{2b} - \frac{a^3}{2b} + \frac{a^3}{6b} \right] \]

The first two terms cancel each other out:

\[ H = R \left[ \frac{a^3}{6b} \right] \] \[ H = \frac{a^3R}{6b} \]

Final Answer Summary

Based on our detailed calculation, the heat dissipated in the wire is \(\frac{a^3R}{6b}\).

Quantity	Formula/Value
Charge \(Q(t)\)	\(at - bt^2\)
Current \(I(t)\)	\(a - 2bt\)
Time interval	\(0\) to \(\frac{a}{2b}\)
Heat Dissipated \(H\)	\(\int I^2 R \, dt\)
Calculated Heat	\(\frac{a^3R}{6b}\)