Question

A RC circuit has an emf of $300\cos (2t)$ volts, a resistance of $150ohms$, a capacitance of $\dfrac{1}{{600}}farad$, and an initial charge on the capacitor of $5C$. Find the charge on the capacitor at any time $t$.
A. $\dfrac{2}{5}(23{e^{ - 4t}} + \sin (2t) + 2\cos (2t))$
B. $\dfrac{1}{5}(23{e^{ - 4t}} + \sin (2t) + 2\cos (2t))$
C. $\dfrac{3}{5}(23{e^{ - 4t}} + \sin (2t) + 2\cos (2t))$
D. $\dfrac{1}{6}(23{e^{ - 4t}} + \sin (2t) + 2\cos (2t))$

Answer 1

Use the circuit equation. Substitute the given values in it and integrate to find the charge. Use the formula of integration by parts to solve it.

Formula used:
$E = \dfrac{{dq}}{{dt}}R + \dfrac{q}{C}$
Where,
$E$ is emf
$\dfrac{{dq}}{{dt}}$ is current
$R$ is the resistance
$q$ is the charge
$C$ is capacitance

Complete step by step answer:
We know that, a circuit equation is given by
$E = \dfrac{{dq}}{{dt}}R + \dfrac{q}{C}$
Where,
$E$ is emf
$\dfrac{{dq}}{{dt}} = i$ is current
$R$ is the resistance
$q$ is the charge
$C$ is capacitance
It is given to us that,
$E = 300\cos (2t)volt$
$R = 150ohm$
$C = \dfrac{1}{{600}}farad$
Substitute the given values in the above equation. We get
$300\cos (2t) = 150\dfrac{{dq}}{{dt}} + 600q$
Rearranging it we can write
$\dfrac{{dq}}{{dt}} + 4q = 2\cos (2t)$ . . . (1)
This is linear equation in terms of $q$ and $t$
Integrating factor:
$IF = {e^{\int {4dt} }} = {e^{4t}}$
The solution of equation (1) can be written as
$q \times IF = \int {IF \times 2\cos (2t)dt}$
$\Rightarrow q{e^{4t}} = 2\int {{e^{4t}}\cos (2t)dt}$ . . . (1)
By using ILATE rule, we can expand the integration using the formula
$\int {I \times IIdx = I\int {IIdx - \int {\left[ {\dfrac{d}{{dx}}I\int {IIdx} } \right]dx} } } + C$
$\Rightarrow q{e^{4t}} = 2\left[ {\cos (2t)\int {{e^{4t}}dt - \int {\left[ {\dfrac{d}{{dt}}\cos (2t)\int {{e^{4t}}dt} } \right]dt} } } \right] + C$
$= 2\left[ {\cos (2t)\dfrac{{{e^{4t}}}}{4} - 2\int { - \sin (2t)} \times \dfrac{{{e^{4t}}}}{4}} \right] + C$
$= 2\left[ {\cos (2t)\dfrac{{{e^{4t}}}}{4} + \dfrac{2}{4}\int {\sin (2t)} \times {e^{4t}}dt} \right] + C$
$= 2\left[ {\cos (2t)\dfrac{{{e^{4t}}}}{4} + \dfrac{1}{2}\left[ {\sin (2t)\dfrac{{{e^{4t}}}}{4} - \int {\dfrac{{\cos (2t)}}{2}\dfrac{{{e^{4t}}}}{4}} dt} \right]} \right] + C$
$= 2\left[ {\cos (2t)\dfrac{{{e^{4t}}}}{4} + \dfrac{1}{8}\sin (2t){e^{4t}} - \dfrac{1}{{16}}\int {\cos (2t){e^{4t}}} dt} \right] + C$
$\Rightarrow q{e^{4t}} = 2\left[ {\cos (2t)\dfrac{{{e^{4t}}}}{4} + \dfrac{1}{8}\sin (2t){e^{4t}} - \dfrac{1}{{16}}q{e^{4t}}} \right] + C$
$\Rightarrow q{e^{4t}} = \dfrac{1}{2}\cos (2t){e^{4t}} + \dfrac{1}{4}\sin (2t){e^{4t}} - \dfrac{1}{8}q{e^{4t}} + C$
$\Rightarrow q{e^{4t}} + \dfrac{1}{8}q{e^{4t}} = \dfrac{1}{2}\cos (2t){e^{4t}} + \dfrac{1}{4}\sin (2t){e^{4t}} + C$
$\Rightarrow \dfrac{9}{8}q{e^{4t}} = \dfrac{1}{2}\cos (2t){e^{4t}} + \dfrac{1}{4}\sin (2t){e^{4t}} + C$ . . . (2)
Let the initial charge be ${q_0} = 5C$ when $t = 0$
Substitute the given values in the above equation. We get
$\Rightarrow \dfrac{9}{8} \times 5{e^0} = \dfrac{1}{2}\cos (0){e^0} + \dfrac{1}{4}\sin (0){e^0} + C$
$\Rightarrow \dfrac{{45}}{8} = \dfrac{1}{2} + C$
Rearranging it we can write
$C = \dfrac{{45}}{8} - \dfrac{1}{2}$
$\Rightarrow C = \dfrac{{45 - 4}}{8}$
$\Rightarrow C = \dfrac{{41}}{8}$
Therefore, equation (2) becomes
$\Rightarrow \dfrac{9}{8}q{e^{4t}} = \dfrac{1}{2}\cos (2t){e^{4t}} + \dfrac{1}{4}\sin (2t){e^{4t}} + \dfrac{{41}}{8}$
Simplifying it, we get
$q = \dfrac{1}{5}(23{e^{ - 4t}} + \sin (2t) + 2\cos (2t))$
Therefore, from the above explanation, the correct answer is, option (B) $\dfrac{1}{5}(23{e^{ - 4t}} + \sin (2t) + 2\cos (2t))$

Note: We integrated equation (1) using the formula called integral by parts. You cannot integrate the product of two functions normally. You need to follow the ILATE rule for that. ILATE is short for, INVERSE, LOGARITHMIC, ALGEBRAIC, TRIGONOMETRIC, EXPONENTIAL function. The first function and the second function is chosen on the basis of the decreasing priority in the word ILATE. Once, you decide the first and the second function, then you can use the formula we discussed in the solution.