The charge flowing through a resistance R varies with time as `Q=2t-8t^(2)` . The total heat produced in the resistance is (for `0letle(1)/(8))`

Options

A. `(R)/(6)` joules
B. `(R)/(3)` joules
C. `(R)/(2)` joules
D. R joules

Answer

To find the total heat produced in the resistance \( R \) when the charge flowing through it varies with time as \( Q = 2t - 8t^2 \), we will follow these steps:

### Step 1: Determine the Current
The current \( I \) is defined as the rate of change of charge with respect to time. Thus, we can express the current as:
\[ I = \frac{dQ}{dt} \]
Given \( Q = 2t - 8t^2 \), we differentiate this with respect to \( t \):
\[ I = \frac{dQ}{dt} = \frac{d}{dt}(2t - 8t^2) = 2 - 16t \]

### Step 2: Calculate the Heat Produced
The total heat \( H \) produced in the resistance \( R \) can be calculated using the formula:
\[ H = \int_0^{t_f} I^2 R \, dt \]
where \( t_f \) is the final time. In this case, \( t_f = \frac{1}{8} \).

Substituting \( I = 2 - 16t \):
\[ H = R \int_0^{\frac{1}{8}} (2 - 16t)^2 \, dt \]

### Step 3: Expand the Expression
Next, we need to expand \( (2 - 16t)^2 \):
\[ (2 - 16t)^2 = 4 - 64t + 256t^2 \]

### Step 4: Integrate the Expression
Now we integrate the expanded expression:
\[ H = R \int_0^{\frac{1}{8}} (4 - 64t + 256t^2) \, dt \]

Calculating the integral term by term:
1. \(\int 4 \, dt = 4t\)
2. \(\int -64t \, dt = -32t^2\)
3. \(\int 256t^2 \, dt = \frac{256}{3}t^3\)

Putting it all together:
\[ H = R \left[ 4t - 32t^2 + \frac{256}{3}t^3 \right]_0^{\frac{1}{8}} \]

### Step 5: Evaluate the Integral at the Limits
Now we evaluate the expression at the limits \( 0 \) and \( \frac{1}{8} \):
\[ H = R \left[ 4\left(\frac{1}{8}\right) - 32\left(\frac{1}{8}\right)^2 + \frac{256}{3}\left(\frac{1}{8}\right)^3 \right] \]
Calculating each term:
1. \( 4\left(\frac{1}{8}\right) = \frac{1}{2} \)
2. \( -32\left(\frac{1}{8}\right)^2 = -32 \cdot \frac{1}{64} = -\frac{1}{2} \)
3. \( \frac{256}{3}\left(\frac{1}{8}\right)^3 = \frac{256}{3} \cdot \frac{1}{512} = \frac{256}{1536} = \frac{1}{6} \)

Now substituting these values:
\[ H = R \left[ \frac{1}{2} - \frac{1}{2} + \frac{1}{6} \right] = R \cdot \frac{1}{6} \]

### Final Result
Thus, the total heat produced in the resistance \( R \) is:
\[ H = \frac{R}{6} \text{ Joules} \]