An amount of charge Q passes through a coil o | vedclass

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Question

(A) The total charge $Q$ passing through the coil is equal to the area under the current-time graph. Since the current decreases uniformly from $I_0$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The total charge $Q$ passing through the coil is equal to the area under the current-time graph. Since the current decreases uniformly from $I_0$ to $0$ in time $T$, the graph is a triangle with base $T$ and height $I_0$.$Q = \frac{1}{2} I_0 T \Rightarrow I_0 = \frac{2 Q}{T}$The current as a function of time is given by $I(t) = I_0 \left(1 - \frac{t}{T}\right) = \frac{2 Q}{T} \left(1 - \frac{t}{T}\right)$.The heat generated $H$ in the coil is given by $H = \int_0^T I^2 R \, dt$.$H = R \int_0^T \left[ \frac{2 Q}{T} \left(1 - \frac{t}{T}\right) \right]^2 \, dt = \frac{4 Q^2 R}{T^2} \int_0^T \left(1 - \frac{t}{T}\right)^2 \, dt$.Let $u = 1 - \frac{t}{T}$, then $du = -\frac{1}{T} dt$, so $dt = -T du$.When $t=0, u=1$; when $t=T, u=0$.$H = \frac{4 Q^2 R}{T^2} \int_1^0 u^2 (-T \, du) = \frac{4 Q^2 R}{T} \int_0^1 u^2 \, du$.$H = \frac{4 Q^2 R}{T} \left[ \frac{u^3}{3} \right]_0^1 = \frac{4 Q^2 R}{3 T}$.

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,

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