The charge flowing through a resistance R var | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given,$Q = at - bt^2$. The current $I$ is given by $I = \frac{dQ}{dt} = a - 2bt$. The current becomes zero when $a - 2bt = 0$,which gives $t = \fr

Vedclass · Accepted Answer

$\frac{a^3 R}{6b}$

Vedclass · Answer

(D) Given,$Q = at - bt^2$. The current $I$ is given by $I = \frac{dQ}{dt} = a - 2bt$. The current becomes zero when $a - 2bt = 0$,which gives $t = \frac{a}{2b}$. The total heat $H$ produced in the resistance $R$ is given by $H = \int_0^{a/2b} I^2 R dt$. Substituting $I = a - 2bt$: $H = R \int_0^{a/2b} (a - 2bt)^2 dt = R \int_0^{a/2b} (a^2 + 4b^2 t^2 - 4abt) dt$. Integrating with respect to $t$: $H = R \left[ a^2 t + \frac{4b^2 t^3}{3} - 2abt^2 \right]_0^{a/2b}$. Substituting the limits: $H = R \left[ a^2 \left( \frac{a}{2b} \right) + \frac{4b^2}{3} \left( \frac{a^3}{8b^3} \right) - 2ab \left( \frac{a^2}{4b^2} \right) \right]$. $H = R \left[ \frac{a^3}{2b} + \frac{a^3}{6b} - \frac{a^3}{2b} \right] = \frac{a^3 R}{6b}$.

The charge flowing through a resistance $R$ varies with time $t$ as $Q = at - bt^2$,where $a$ and $b$ are positive constants. The total heat produced in $R$ is

Similar Questions

What is the equivalent resistance of the circuit in $\Omega$?

The potential difference between points $A$ and $B$ of the adjoining figure is

Find the potential of $J$ with respect to $G$ in $V$.

In the circuit shown here, what is the value of the unknown resistor $R$ so that the total resistance of the circuit between points $P$ and $Q$ is also equal to $R$?

In the circuit shown,the reading of the ammeter (ideal) is the same with both switches open as with both closed. Find the value of resistance $R$ in $\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module