The charge supplied by a source varies with t | vedclass

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(B) The total current $I$ supplied by the source is given by $I = \frac{dQ}{dt} = \frac{d}{dt}(at - bt^2) = a - 2bt$.The current becomes zero when $a

Vedclass · Accepted Answer

$\frac{a^3R}{27b}$

Vedclass · Answer

(B) The total current $I$ supplied by the source is given by $I = \frac{dQ}{dt} = \frac{d}{dt}(at - bt^2) = a - 2bt$.The current becomes zero when $a - 2bt = 0$, which gives $t = \frac{a}{2b}$.The resistors $R$ and $2R$ are connected in parallel. Using the current divider rule, the current $I'$ flowing through the $2R$ resistor is:$I' = I 	imes \frac{R}{R + 2R} = \frac{1}{3}I = \frac{1}{3}(a - 2bt)$.The heat $H$ produced in the $2R$ resistor is given by $H = \int_0^{a/2b} (I')^2 (2R) dt$.$H = \int_0^{a/2b} \left[ \frac{1}{3}(a - 2bt) ight]^2 (2R) dt = \frac{2R}{9} \int_0^{a/2b} (a - 2bt)^2 dt$.Let $u = a - 2bt$, then $du = -2b dt$, or $dt = -\frac{du}{2b}$.When $t=0, u=a$. When $t=a/2b, u=0$.$H = \frac{2R}{9} \int_a^0 u^2 \left( -\frac{du}{2b} ight) = \frac{2R}{9(2b)} \int_0^a u^2 du = \frac{R}{9b} \left[ \frac{u^3}{3} ight]_0^a = \frac{R}{9b} \left( \frac{a^3}{3} ight) = \frac{a^3R}{27b}$.

Order	$V \ (V)$	$I \ (A)$
$1$	$1.0$	$0.08$
$2$	$0.5$	$0.18$
$3$	$0.8$	$0.12$

The charge supplied by a source varies with time $t$ as $Q = at - bt^2$. Find the total heat produced in the $2R$ resistor before the current reduces to zero.

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