A total charge Q flows across a resistor R during a time interval T in such a way that the current versus time graph for 0 to T is like the loop of a sine curve in the range `0 "to" pi`. The total heat generated in the resistor is

Options

A. `Q^2pi^2R//8T`
B. `2Q^2pi^2R//T`
C. `Q^2pi^2R//T`
D. `Q^2pi^2R//2T`

Explanation

To find the total heat generated in the resistor when a total charge \( Q \) flows across it during a time interval \( T \), we can follow these steps:

### Step 1: Understand the Current Function
The current \( I(t) \) is given by the equation of a sine wave:
\[ I(t) = I_0 \sin\left(\frac{\pi t}{T}\right) \]
where \( I_0 \) is the maximum current.

### Step 2: Relate Charge and Current
The total charge \( Q \) that flows through the resistor can be expressed as:
\[ Q = \int_0^T I(t) \, dt \]
Substituting the expression for \( I(t) \):
\[ Q = \int_0^T I_0 \sin\left(\frac{\pi t}{T}\right) \, dt \]

### Step 3: Calculate the Integral
To solve the integral:
\[ Q = I_0 \int_0^T \sin\left(\frac{\pi t}{T}\right) \, dt \]
Using the substitution \( u = \frac{\pi t}{T} \) gives \( dt = \frac{T}{\pi} du \). The limits change from \( 0 \) to \( \pi \):
\[ Q = I_0 \cdot \frac{T}{\pi} \int_0^{\pi} \sin(u) \, du \]
The integral of \( \sin(u) \) from \( 0 \) to \( \pi \) is \( 2 \):
\[ Q = I_0 \cdot \frac{T}{\pi} \cdot 2 = \frac{2 I_0 T}{\pi} \]

### Step 4: Solve for \( I_0 \)
Rearranging gives:
\[ I_0 = \frac{Q \pi}{2T} \]

### Step 5: Calculate the Heat Generated
The heat \( H \) generated in the resistor can be calculated using:
\[ H = \int_0^T I^2(t) R \, dt \]
Substituting \( I(t) \):
\[ H = R \int_0^T \left(I_0 \sin\left(\frac{\pi t}{T}\right)\right)^2 \, dt \]
This simplifies to:
\[ H = R I_0^2 \int_0^T \sin^2\left(\frac{\pi t}{T}\right) \, dt \]

### Step 6: Evaluate the Integral
Using the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):
\[ H = R I_0^2 \cdot \frac{T}{2} \left(1 - \frac{1}{2}\int_0^T \cos\left(\frac{2\pi t}{T}\right) \, dt\right) \]
The integral of \( \cos\left(\frac{2\pi t}{T}\right) \) over one complete cycle is zero:
\[ H = R I_0^2 \cdot \frac{T}{2} \]

### Step 7: Substitute \( I_0 \)
Substituting \( I_0 = \frac{Q \pi}{2T} \):
\[ H = R \left(\frac{Q \pi}{2T}\right)^2 \cdot \frac{T}{2} \]
This simplifies to:
\[ H = R \cdot \frac{Q^2 \pi^2}{4T^2} \cdot \frac{T}{2} = \frac{Q^2 \pi^2 R}{8T} \]

### Final Answer
The total heat generated in the resistor is:
\[ H = \frac{Q^2 \pi^2 R}{8T} \]