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. \(Q^2RT\)

Hint:

The correct option is B ^I₀2RT/₃
Area under the I-t graph represents charge.
Therefore, ¹/₂ × T × I₀ = Q → I₀ = ^2Q/_T
H = ∫ I²Rdt = 2 ∫^T₀ ^2I₀t/_T Rdt
H = ^I₀2RT/₃

Answer 1

The Correct Option is A

The correct answer is option (A): \(\frac{4Q^2R}{3T}\)
The corresponding 
i−t graph will be a straight line with 
i decreasing from a peak value 
(say i)to zero in time t.
∴\(i=i-(\frac{i}{t})t\)
(y=−mx+c)……(i)
\(i=\frac{2q}{t}\)
Now, at time t, heat produced in a short interval dt is,
dH=i²Rdt
\(\Rightarrow dH=(\frac{2q}{t}-\frac{2qt^2}{t^2})Rdt\)
∴ Total heat produced,
\(\int_{0}^{H}dH=\int_{0}^{t}(\frac{2q}{t}-\frac{2qt^2}{t^2})Rdt\)
\(\Rightarrow H=\frac{4q^2R}{3t}\)