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 will analyze the situation of charge flowing through a coil of resistance \( R \) and how it relates to the heat generated in the coil.

### Step 1: Understanding the Charge Flow
We are given that the charge flowing through the coil reduces by a factor of \( \frac{1}{2} \) every time interval \( t_0 \). This indicates that the charge flow follows an exponential decay pattern, similar to radioactive decay.

### Step 2: Expressing Current as a Function of Time
Since the charge reduces by half every \( t_0 \), we can express the current \( I(t) \) as:
\[ I(t) = I_0 e^{-\lambda t} \]
where \( \lambda = \frac{\ln 2}{t_0} \) and \( I_0 \) is the initial current.

### Step 3: Relating 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: Evaluating the Integral for Charge
Evaluating the integral:
\[ Q = I_0 \int_0^{\infty} e^{-\lambda t} \, dt = I_0 \left[ -\frac{1}{\lambda} e^{-\lambda t} \right]_0^{\infty} = I_0 \left( 0 + \frac{1}{\lambda} \right) = \frac{I_0}{\lambda} \]
Thus, we have:
\[ I_0 = Q \lambda \]

### Step 5: Heat Generated in the Coil
The heat generated \( H \) in the coil can be calculated using:
\[ H = \int_0^{\infty} I^2 R \, dt \]
Substituting \( I(t) \):
\[ H = \int_0^{\infty} (I_0 e^{-\lambda t})^2 R \, dt = R I_0^2 \int_0^{\infty} e^{-2\lambda t} \, dt \]

### Step 6: Evaluating the Integral for Heat
Evaluating the integral:
\[ H = R I_0^2 \left[ -\frac{1}{2\lambda} e^{-2\lambda t} \right]_0^{\infty} = R I_0^2 \left( 0 + \frac{1}{2\lambda} \right) = \frac{R I_0^2}{2\lambda} \]

### Step 7: Substituting \( I_0 \) in Terms of \( Q \)
Now substituting \( I_0 = Q \lambda \):
\[ H = \frac{R (Q \lambda)^2}{2\lambda} = \frac{Q^2 \lambda R}{2} \]

### Step 8: Replacing \( \lambda \)
Since \( \lambda = \frac{\ln 2}{t_0} \):
\[ H = \frac{Q^2 R \ln 2}{2 t_0} \]

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