The charge flowing through a resistance `R` varies with time t as `Q = at - bt^(2)`. The total heat produced in `R` is

Options

A. `(a^(3) R)/(6 b)`
B. `(a^(3) R)/(3 b)`
C. `(a^(3) R)/(2 b)`
D. `(a^(3) R)/( b)`

Explanation

To find the total heat produced in a resistance \( R \) when the charge flowing through it varies with time as \( Q = at - bt^2 \), we can follow these steps:

### Step 1: Find the current \( I \)
The current \( I \) is defined as the rate of change of charge with respect to time. Therefore, we differentiate \( Q \) with respect to \( t \):

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

Calculating the derivative:

\[ I = a - 2bt \]

### Step 2: Determine the limits of time
The current \( I \) will be zero at two points: when \( t = 0 \) and when \( I = 0 \):

\[ a - 2bt = 0 \implies t = \frac{a}{2b} \]

Thus, the current flows from \( t = 0 \) to \( t = t_n = \frac{a}{2b} \).

### Step 3: Calculate the heat produced
The heat \( dH \) produced in a resistance \( R \) during a small time interval \( dt \) is given by:

\[ dH = I^2 R \, dt \]

Substituting the expression for \( I \):

\[ dH = (a - 2bt)^2 R \, dt \]

### Step 4: Integrate to find total heat
Now, we need to integrate \( dH \) from \( t = 0 \) to \( t = t_n \):

\[ H = \int_0^{t_n} (a - 2bt)^2 R \, dt \]

Expanding \( (a - 2bt)^2 \):

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

Thus, the integral becomes:

\[ H = R \int_0^{t_n} (a^2 - 4abt + 4b^2t^2) \, dt \]

### Step 5: Evaluate the integral
Now, we can evaluate the integral term by term:

1. \(\int_0^{t_n} a^2 \, dt = a^2 t_n\)
2. \(\int_0^{t_n} -4abt \, dt = -4ab \frac{t_n^2}{2} = -2abt_n^2\)
3. \(\int_0^{t_n} 4b^2t^2 \, dt = 4b^2 \frac{t_n^3}{3}\)

Substituting \( t_n = \frac{a}{2b} \):

1. \( a^2 t_n = a^2 \left(\frac{a}{2b}\right) = \frac{a^3}{2b} \)
2. \( -2ab t_n^2 = -2ab \left(\frac{a^2}{4b^2}\right) = -\frac{a^3}{2b} \)
3. \( 4b^2 t_n^3 = 4b^2 \left(\frac{a^3}{8b^3}\right) = \frac{a^3}{2b} \)

### Step 6: Combine the results
Now, substituting back into the expression for \( H \):

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

### Final Answer
Thus, the total heat produced in the resistance \( R \) is:

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