Question

An amount of charge Q passes through a coil of resistance R.If the current in the coil decreases to zero at a uniform rate during time T, then the amount of heat generated in the coil will be,

• A. \(\frac{4Q^2R}{3T}\)
• B. \(\frac{2QR}{3T}\)
• C. \(\frac{Q^2T}{4R}\)
• D. \(Q^2RT\)

Answer 1

The Correct Option is A

Step 1: Expression for Current as a Function of Time

The current $ I(t) $ in the coil decreases uniformly from an initial value $ I_0 $ to zero over time $ T $. This implies that the current decreases linearly with time:

$$ I(t) = I_0 \left( 1 - \frac{t}{T} \right) $$

At $ t = 0 $, $ I(t) = I_0 $, and at $ t = T $, $ I(t) = 0 $.

Step 2: Relationship Between Charge and Current

The total charge $ Q $ passing through the coil is related to the current by:

$$ Q = \int_{0}^{T} I(t) \, dt $$

Substitute $ I(t) = I_0 \left( 1 - \frac{t}{T} \right) $:

$$ Q = \int_{0}^{T} I_0 \left( 1 - \frac{t}{T} \right) \, dt $$

Split the integral:

$$ Q = I_0 \int_{0}^{T} 1 \, dt - I_0 \int_{0}^{T} \frac{t}{T} \, dt $$

Evaluate each term:

$$ \int_{0}^{T} 1 \, dt = T $$

$$ \int_{0}^{T} \frac{t}{T} \, dt = \frac{1}{T} \int_{0}^{T} t \, dt = \frac{1}{T} \left[ \frac{t^2}{2} \right]_0^T = \frac{T}{2} $$

Thus:

$$ Q = I_0 T - I_0 \frac{T}{2} = I_0 \frac{T}{2} $$

Rearrange to find $ I_0 $:

$$ I_0 = \frac{2Q}{T} $$

Step 3: Heat Generated in the Coil

The heat generated in the coil is given by:

$$ H = \int_{0}^{T} I^2(t) R \, dt $$

Substitute $ I(t) = I_0 \left( 1 - \frac{t}{T} \right) $:

$$ H = \int_{0}^{T} \left[ I_0 \left( 1 - \frac{t}{T} \right) \right]^2 R \, dt $$

Expand the square:

$$ H = I_0^2 R \int_{0}^{T} \left( 1 - \frac{2t}{T} + \frac{t^2}{T^2} \right) \, dt $$

Split the integral:

$$ H = I_0^2 R \left[ \int_{0}^{T} 1 \, dt - \int_{0}^{T} \frac{2t}{T} \, dt + \int_{0}^{T} \frac{t^2}{T^2} \, dt \right] $$

Evaluate each term:

$$ \int_{0}^{T} 1 \, dt = T $$

$$ \int_{0}^{T} \frac{2t}{T} \, dt = \frac{2}{T} \int_{0}^{T} t \, dt = \frac{2}{T} \left[ \frac{t^2}{2} \right]_0^T = \frac{2}{T} \cdot \frac{T^2}{2} = T $$

$$ \int_{0}^{T} \frac{t^2}{T^2} \, dt = \frac{1}{T^2} \int_{0}^{T} t^2 \, dt = \frac{1}{T^2} \left[ \frac{t^3}{3} \right]_0^T = \frac{1}{T^2} \cdot \frac{T^3}{3} = \frac{T}{3} $$

Substitute back:

$$ H = I_0^2 R \left[ T - T + \frac{T}{3} \right] = I_0^2 R \cdot \frac{T}{3} $$

Substitute $ I_0 = \frac{2Q}{T} $:

$$ H = \left( \frac{2Q}{T} \right)^2 R \cdot \frac{T}{3} $$

Simplify:

$$ H = \frac{4Q^2}{T^2} R \cdot \frac{T}{3} = \frac{4Q^2R}{3T} $$

Step 4: Final Answer

The correct option is:

$$ \boxed{\frac{4Q^2R}{3T}} $$

Answer 2

The correct answer is option (A): \(\frac{4Q^2R}{3T}\) 

Explanation:

The corresponding i–t graph is a straight line where current i decreases linearly from a peak value to zero in time t.

So, the current at time t is given by the equation of a straight line:
\(i = i - \left( \frac{i}{t} \right) t\)
which resembles the equation y = −mx + c.

Now, the average current is:
\(i_{\text{avg}} = \frac{2q}{t}\)

At any instant, the heat produced in a short time dt is:
\(dH = i^2 R \, dt\)

Substituting the value of i(t) into the expression for heat:
\(dH = \left( \frac{2q}{t} - \frac{2q t^2}{t^2} \right)^2 R \, dt\)

Now, total heat produced over time t is:

\(H = \int_{0}^{t} \left( \frac{2q}{t} - \frac{2q t}{t^2} \right)^2 R \, dt\)

On solving the integral, we get:

\(H = \frac{4q^2 R}{3t}\)