Through a coil of resistance R charge is flowing in such a manner that in every time interval `t_(0)` the amount of charge flow is reduced by a factor of `1//2`. If a total of q charge flows through the coil in very long time, then the total amount of heat generated in the coil will be

Options

A. `(1)/(4t_(0))q^(2)Rlog_(e)2`
B. `(3)/(2)(q^(2)Rlog_(e)2)/(t_(0))`
C. `(q^(2)Rlog_(e)2)/(2t_(0))`
D. `(3q^(2)Rlog_(e)2)/(4t_(0))`

Answer

To solve the problem step by step, we need to analyze the situation where the charge flowing through the coil decreases by a factor of 1/2 every time interval \( t_0 \). We want to find the total heat generated in the coil when a total charge \( Q \) flows through it over a long time.

### Step 1: Understand the current decay
Given that the amount of charge flowing decreases by a factor of \( \frac{1}{2} \) every interval \( t_0 \), we can express the current \( I(t) \) as an exponentially decaying function:
\[ I(t) = I_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

### Step 2: Determine the decay constant
Since the charge decreases by half every \( t_0 \), we can relate \( \lambda \) to \( t_0 \):
\[ \lambda = \frac{\ln(2)}{t_0} \]

### Step 3: Relate charge and current
The total charge \( Q \) flowing through the coil can be expressed as:
\[ Q = \int_0^\infty I(t) \, dt \]
Substituting the expression for \( I(t) \):
\[ Q = \int_0^\infty I_0 e^{-\lambda t} \, dt \]

### Step 4: Evaluate the integral
The integral of \( e^{-\lambda t} \) from 0 to infinity is:
\[ \int_0^\infty e^{-\lambda t} \, dt = \frac{1}{\lambda} \]
Thus, we have:
\[ Q = I_0 \cdot \frac{1}{\lambda} \]
Rearranging gives:
\[ I_0 = \lambda Q \]

### Step 5: Calculate the heat generated
The heat generated \( H \) in the coil can be expressed as:
\[ H = \int_0^\infty I^2(t) R \, dt \]
Substituting \( I(t) \):
\[ H = \int_0^\infty (I_0 e^{-\lambda t})^2 R \, dt = R I_0^2 \int_0^\infty e^{-2\lambda t} \, dt \]
The integral \( \int_0^\infty e^{-2\lambda t} \, dt \) evaluates to:
\[ \int_0^\infty e^{-2\lambda t} \, dt = \frac{1}{2\lambda} \]
Thus, we have:
\[ H = R I_0^2 \cdot \frac{1}{2\lambda} \]

### Step 6: Substitute \( I_0 \) into the heat equation
Now substituting \( I_0 = \lambda Q \):
\[ H = R (\lambda Q)^2 \cdot \frac{1}{2\lambda} = \frac{R \lambda Q^2}{2} \]

### Step 7: Substitute \( \lambda \)
Finally, substituting \( \lambda = \frac{\ln(2)}{t_0} \):
\[ H = \frac{R Q^2 \ln(2)}{2 t_0} \]

### Final Result
Thus, the total amount of heat generated in the coil is:
\[ H = \frac{Q^2 R \ln(2)}{2 t_0} \]