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

Answer

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

### Step 1: Differentiate Charge to Find Current
The current \( I \) is defined as the rate of change of charge with respect to time. Therefore, we can express the current as:
\[ I = \frac{dQ}{dt} = \frac{d}{dt}(at - bt^2) \]
Calculating the derivative:
\[ I = a - 2bt \]

### Step 2: Identify the Time When Current is Zero
To find the time when the current is zero, we set \( I = 0 \):
\[ a - 2bt = 0 \]
Solving for \( t \):
\[ t = \frac{a}{2b} \]

### Step 3: Write the Expression for Heat Produced
The heat produced \( H \) in a resistor is given by the formula:
\[ H = \int_0^T I^2 R \, dt \]
where \( T \) is the time interval from \( 0 \) to \( \frac{a}{2b} \).

### Step 4: Substitute Current into the Heat Formula
Substituting \( I = a - 2bt \) into the heat formula:
\[ H = \int_0^{\frac{a}{2b}} (a - 2bt)^2 R \, dt \]

### Step 5: Expand the Expression
Expanding \( (a - 2bt)^2 \):
\[ (a - 2bt)^2 = a^2 - 4abt + 4b^2t^2 \]
Thus, the heat produced becomes:
\[ H = R \int_0^{\frac{a}{2b}} (a^2 - 4abt + 4b^2t^2) \, dt \]

### Step 6: Integrate Each Term
Now, we integrate each term:
\[ H = R \left[ a^2t - 2abt^2 + \frac{4b^2t^3}{3} \right]_0^{\frac{a}{2b}} \]
Evaluating at the limits:
\[ H = R \left[ a^2 \left(\frac{a}{2b}\right) - 2ab \left(\frac{a}{2b}\right)^2 + \frac{4b^2 \left(\frac{a}{2b}\right)^3}{3} \right] \]

### Step 7: Simplify the Expression
Calculating each term:
1. \( a^2 \left(\frac{a}{2b}\right) = \frac{a^3}{2b} \)
2. \( -2ab \left(\frac{a}{2b}\right)^2 = -2ab \cdot \frac{a^2}{4b^2} = -\frac{a^3}{2b} \)
3. \( \frac{4b^2 \left(\frac{a}{2b}\right)^3}{3} = \frac{4b^2 \cdot \frac{a^3}{8b^3}}{3} = \frac{a^3}{6b} \)

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

### Final Answer
Thus, the total heat produced in the resistance \( R \) is:
\[ H = \frac{a^3 R}{6} \]