A total charge Q flows across a resistor R du | vedclass

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

Question

(A) The current follows the form $I(t) = I_0 \sin(\omega t)$. Since the loop is for $0 \rightarrow T$,we have $\omega T = \pi$,so $\omega = \pi/T$.The

Vedclass · Accepted Answer

$Q^2\pi^2R / 8T$

Vedclass · Answer

(A) The current follows the form $I(t) = I_0 \sin(\omega t)$. Since the loop is for $0 \rightarrow T$,we have $\omega T = \pi$,so $\omega = \pi/T$.The total charge $Q$ is the integral of current over time:$Q = \int_0^T I_0 \sin(\frac{\pi}{T} t) dt = I_0 [-\frac{T}{\pi} \cos(\frac{\pi}{T} t)]_0^T = I_0 \frac{T}{\pi} (1 - (-1)) = \frac{2 I_0 T}{\pi}$.Solving for $I_0$,we get $I_0 = \frac{Q \pi}{2T}$.The heat generated $H$ is given by $\int_0^T I^2 R dt$:$H = R \int_0^T I_0^2 \sin^2(\frac{\pi}{T} t) dt = R I_0^2 \int_0^T \frac{1 - \cos(\frac{2\pi}{T} t)}{2} dt$.$H = \frac{R I_0^2}{2} [t - \frac{T}{2\pi} \sin(\frac{2\pi}{T} t)]_0^T = \frac{R I_0^2}{2} [T - 0] = \frac{R I_0^2 T}{2}$.Substituting $I_0 = \frac{Q \pi}{2T}$:$H = \frac{R T}{2} (\frac{Q \pi}{2T})^2 = \frac{R T Q^2 \pi^2}{2 \cdot 4 T^2} = \frac{Q^2 \pi^2 R}{8 T}$.

$A$ total charge $Q$ flows across a resistor $R$ during a time interval $T$ in such a way that the current vs. time graph for $0 \rightarrow T$ is like the loop of a sine curve in the range $0 \rightarrow \pi$. The total heat generated in the resistor is

Similar Questions

$A$ constant current $i$ is passed through a resistor. Taking the temperature coefficient of resistance into account,indicate which of the plots shown in the figure best represents the rate of production of thermal energy in the resistor.

The battery of a mobile phone is rated as $4.2 \ V, 5800 \ mAh$. How much energy is stored in it when fully charged (in $kJ$)?

From where does electrical power derive? Explain it.

If the power in the external resistance $R$ is maximum,then which of the following statements are correct?
$(i) R = r$
$(ii)$ Power in $R$ is $\frac{E^2}{4R}$
$(iii)$ Input power is $\frac{E^2}{2R}$
$(iv)$ Efficiency is $50\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module