Chapter 1, Problem 3P

Find the charge q(t) flowing through a device if the current is:

1. (a) i(t) = 3 A, q(0) = 1 C
2. (b) i(t) = (2t + 5) mA, q(0) = 0
3. (c) i(t) = 20 cos(10t + π/6) μA, q(0) = 2μC
4. (d) i(t) = 10e^‒30t sin 40t A, q (0) = 0

To determine

Calculate the charge $q\left(t\right)$ flowing through a device when the current is $i\left(t\right)=3\text{A}$, and $q\left(0\right)=1\text{C}$.

Answer to Problem 3P

The charge $q\left(t\right)$ flowing through the device is $\underset{\_}{\left(3t+1\right)\text{C}}$.

Explanation of Solution

Given data:

The current flowing through the device $\left[i\left(t\right)\right]$ is 3 A.

The charge at initial state $\left[q\left(0\right)\right]$ is 1 C.

Formula used:

Write the expression for charge entering to the device as follows:

$q\left(t\right)={\displaystyle \int i\left(t\right)dt}+q\left(0\right)$ (1)

Here,

$i\left(t\right)$ is the current through the device, and

$q\left(0\right)$ is the charge at initial state.

Calculation:

Substitute 3 A for $i\left(t\right)$ and 1 C for $q\left(0\right)$ in equation (1) to obtain the charge flowing through the device as follows:

$\begin{array}{c}q\left(t\right)={\displaystyle \int \left(3\text{A}\right)dt}+1\text{C}\\ =3t\text{C}+1\text{C}\\ =\left(3t+1\right)\text{C}\end{array}$

Conclusion:

Thus, the charge $q\left(t\right)$ flowing through the device is $\underset{\_}{\left(3t+1\right)\text{C}}$.

To determine

Calculate the charge $q\left(t\right)$ flowing through a device when the current is $i\left(t\right)=\left(2t+5\right)\text{mA}$, and $q\left(0\right)=0\text{mC}$.

Answer to Problem 3P

The charge $q\left(t\right)$ flowing through the device is $\underset{\_}{\left(t^2+5t\right)\text{mC}}$.

Explanation of Solution

Given data:

The current flowing through the device $\left[i\left(t\right)\right]$ is $\left(2t+5\right)\text{mA}$.

The charge at initial state $\left[q\left(0\right)\right]$ is $0\text{mC}$.

Calculation:

Substitute $\left(2t+5\right)\text{mA}$ for $i\left(t\right)$ and $0\text{mC}$ for $q\left(0\right)$ in equation (1) to obtain the charge flowing through the device as follows:

$\begin{array}{c}q\left(t\right)={\displaystyle \int \left[\left(2t+5\right)\text{mA}\right]dt}+0\text{mC}\\ ={\displaystyle \int \left[\left(2t\right)\text{mA}\right]dt}+{\displaystyle \int \left[\left(5\right)\text{mA}\right]dt}\\ =2\left(\frac{t^2}{2}\right)\text{mC}+5t\text{mC}\\ =\left(t^2+5t\right)\text{mC}\end{array}$

Conclusion:

Thus, the charge $q\left(t\right)$ flowing through the device is $\underset{\_}{\left(t^2+5t\right)\text{mC}}$.

To determine

Calculate the charge $q\left(t\right)$ flowing through a device when the current is $i\left(t\right)=20\cos \left(10t+\pi /6\right)\mu \text{A}$, and $q\left(0\right)=2\mu \text{C}$.

Answer to Problem 3P

The charge $q\left(t\right)$ flowing through the device is $\underset{\_}{2\left[\sin \left(10t+\pi /6\right)+1\right]\mu \text{C}}$.

Explanation of Solution

Given data:

The current flowing through the device $\left[i\left(t\right)\right]$ is $20\cos \left(10t+\pi /6\right)\mu \text{A}$.

The charge at initial state $\left[q\left(0\right)\right]$ is $2\mu \text{C}$.

Calculation:

Substitute $20\cos \left(10t+\pi /6\right)\mu \text{A}$ for $i\left(t\right)$ and $2\mu \text{C}$ for $q\left(0\right)$ in equation (1) to obtain the charge flowing through the device as follows:

$\begin{array}{c}q\left(t\right)={\displaystyle \int \left[20\cos \left(10t+\pi /6\right)\mu \text{A}\right]dt}+2\mu \text{C}\\ =\frac{20\sin \left(10t+\pi /6\right)}{10}\mu \text{C}+2\mu \text{C}\left\{\because {\displaystyle \int \cos \left(at\right)dt}=\frac{\sin \left(at\right)}{a}\right\}\\ =2\sin \left(10t+\pi /6\right)\mu \text{C}+2\mu \text{C}\\ =2\left[\sin \left(10t+\pi /6\right)+1\right]\mu \text{C}\end{array}$

Conclusion:

Thus, the charge $q\left(t\right)$ flowing through the device is $\underset{\_}{2\left[\sin \left(10t+\pi /6\right)+1\right]\mu \text{C}}$.

To determine

Calculate the charge $q\left(t\right)$ flowing through a device when the current is $i\left(t\right)=10e^{-30t}\sin \left(40t\right)\text{A}$, and $q\left(0\right)=0\text{C}$.

Answer to Problem 3P

The cha charge $q\left(t\right)$ flowing through the device is $\underset{\_}{-e^{-30t}\left[0.12\sin \left(40t\right)+0.16\cos \left(40t\right)\right]\text{C}}$.

Explanation of Solution

Given data:

The current flowing through the device $\left[i\left(t\right)\right]$ is $10e^{-30t}\sin \left(40t\right)\text{A}$.

The charge at initial state $\left[q\left(0\right)\right]$ is $0\text{C}$.

Calculation:

Substitute $10e^{-30t}\sin \left(40t\right)\text{A}$ for $i\left(t\right)$ and 0 C for $q\left(0\right)$ in equation (1) to obtain the charge flowing through the device as follows:

$\begin{array}{c}q\left(t\right)={\displaystyle \int \left[10e^{-30t}\sin \left(40t\right)\text{A}\right]dt}+0\text{C}\\ =10\left\{{\displaystyle \int \left[e^{-30t}\sin \left(40t\right)\right]dt}\right\}\text{C}\\ =10\left\{\frac{e^{-30t}\left[\left(-30\right)\sin \left(40t\right)-40\cos \left(40t\right)\right]}{{\left(-30\right)}^2+{40}^2}+C\right\}\text{C}\left\{\begin{array}{l}\because {\displaystyle \int e^{at}\sin \left(bt\right)dt}=\\ \frac{e^{at}\left[a\sin \left(bt\right)-b\cos \left(bt\right)\right]}{a^2+b^2}+C\end{array}\right\}\\ =10\left\{\frac{e^{-30t}\left[-30\sin \left(40t\right)-40\cos \left(40t\right)\right]}{900+1600}\right\}\text{C}\end{array}$

Simplify the expression as follows:

$\begin{array}{c}q\left(t\right)=e^{-30t}\left[\frac{-300\sin \left(40t\right)}{2500}-\frac{400\cos \left(40t\right)}{2500}\right]\text{C}\\ =-e^{-30t}\left[0.12\sin \left(40t\right)+0.16\cos \left(40t\right)\right]\text{C}\end{array}$

Conclusion:

Thus, the charge $q\left(t\right)$ flowing through the device is $\underset{\_}{-e^{-30t}\left[0.12\sin \left(40t\right)+0.16\cos \left(40t\right)\right]\text{C}}$.