Chapter 1, Problem 2P

Determine the current flowing through an element if the charge flow is given by

1. (a) q(t) = (3) mC
2. (b) q(t) = (4t² + 20t ‒ 4) C
3. (c) q(t) = (15e^−3t − 2e^−18t) nC
4. (d) q(t) = 5t²(3t³ + 4) pC
5. (e) q(t) = 2e^−3t sin (20πt) μC

To determine

Calculate the current flowing through an element when the charge flow is $q\left(t\right)=3\text{mC}$.

Answer to Problem 2P

The current flowing through an element when the charge flow is $q\left(t\right)=3\text{mC}$ is $\underset{\_}{0\text{mA}}$.

Explanation of Solution

Given data:

The charge flow $\left[q\left(t\right)\right]$ is $3\text{mC}$.

Formula used:

Write the expression for current as follows:

$i\left(t\right)=\frac{dq\left(t\right)}{dt}$ (1)

Here,

$q\left(t\right)$ is the charge flow through the element.

Calculation:

Substitute $3\text{mC}$ for $q\left(t\right)$ in equation (1) to obtain the current flowing through the element as follows:

$\begin{array}{c}i\left(t\right)=\frac{d\left(3\text{mC}\right)}{dt}\\ =\frac{d}{dt}\left(3\times {10}^{-3}\text{C}\right)\\ =0\text{mA}\left\{\because \frac{d}{dt}\left(\text{Constant}\right)=0\right\}\end{array}$

Conclusion:

Thus, the current flowing through an element when the charge flow is $q\left(t\right)=3\text{mC}$ is $\underset{\_}{0\text{mA}}$.

To determine

Calculate the current flowing through an element when the charge flow is $q\left(t\right)=\left(4t^2+20t-4\right)\text{C}$.

Answer to Problem 2P

The current flowing through an element when the charge flow of $q\left(t\right)=\left(4t^2+20t-4\right)\text{C}$ is $\underset{\_}{\left(8t+20\right)\text{A}}$.

Explanation of Solution

Given data:

The charge flow $\left[q\left(t\right)\right]$ is $\left(4t^2+20t-4\right)\text{C}$.

Calculation:

Substitute $\left(4t^2+20t-4\right)\text{C}$ for $q\left(t\right)$ in equation (1) to obtain the current flowing through the element as follows:

$\begin{array}{c}i\left(t\right)=\frac{d\left[\left(4t^2+20t-4\right)\text{C}\right]}{dt}\\ =\left[\frac{d}{dt}\left(4t^2\right)+\frac{d}{dt}\left(20t\right)-\frac{d}{dt}\left(4\right)\right]\text{A}\\ =\left[\left(4\right)\left(2t\right)+\left(20\right)\left(1\right)-0\right]\text{A}\\ =\left(8t+20\right)\text{A}\end{array}$

Conclusion:

Thus, the current flowing through an element when the charge flow of $q\left(t\right)=\left(4t^2+20t-4\right)\text{C}$ is $\underset{\_}{\left(8t+20\right)\text{A}}$.

To determine

Calculate the current flowing through an element when the charge flow is $q\left(t\right)=\left(15e^{-3t}-2e^{-18t}\right)\text{nC}$.

Answer to Problem 2P

The current flowing through an element when the charge flow of $q\left(t\right)=\left(15e^{-3t}-2e^{-18t}\right)\text{nC}$ is $\underset{\_}{\left(-45e^{-3t}+36e^{-18t}\right)\text{nA}}$.

Explanation of Solution

Given data:

The charge flow $\left[q\left(t\right)\right]$ is $\left(15e^{-3t}-2e^{-18t}\right)\text{nC}$.

Calculation:

Substitute $\left(15e^{-3t}-2e^{-18t}\right)\text{nC}$ for $q\left(t\right)$ in equation (1) to obtain the current flowing through the element as follows:

$\begin{array}{c}i\left(t\right)=\frac{d\left[\left(15e^{-3t}-2e^{-18t}\right)\text{nC}\right]}{dt}\\ =\left[\frac{d}{dt}\left(15e^{-3t}\right)-\frac{d}{dt}\left(2e^{-18t}\right)\right]\text{nA}\\ =\left[\left(15\right)\left(-3\right)e^{-3t}-\left(2\right)\left(-18\right)e^{-18t}\right]\text{nA}\left\{\because \frac{d}{dt}{e}^{at}=a{e}^{at}\right\}\\ =\left(-45e^{-3t}+36e^{-18t}\right)\text{nA}\end{array}$

Conclusion:

Thus, the current flowing through an element when the charge flow of $q\left(t\right)=\left(15e^{-3t}-2e^{-18t}\right)\text{nC}$ is $\underset{\_}{\left(-45e^{-3t}+36e^{-18t}\right)\text{nA}}$.

To determine

Calculate the current flowing through an element when the charge flow is $q\left(t\right)=5t^2\left(3t^3+4\right)\text{pC}$.

Answer to Problem 2P

The current flowing through an element when the charge flow of $q\left(t\right)=5t^2\left(3t^3+4\right)\text{pC}$ is $\underset{\_}{\left(75t^4+40t\right)\text{pA}}$.

Explanation of Solution

Given data:

The charge flow $\left[q\left(t\right)\right]$ is $5t^2\left(3t^3+4\right)\text{pC}$.

Calculation:

Substitute $5t^2\left(3t^3+4\right)\text{pC}$ for $q\left(t\right)$ in equation (1) to obtain the current flowing through the element as follows:

$\begin{array}{c}i\left(t\right)=\frac{d\left[5t^2\left(3t^3+4\right)\text{pC}\right]}{dt}\\ =\left[\frac{d}{dt}\left(15t^5\right)+\frac{d}{dt}\left(20t^2\right)\right]\text{pA}\\ =\left[\left(15\right)\left(5t^4\right)+\left(20\right)\left(2t\right)\right]\text{pA}\\ =\left(75t^4+40t\right)\text{pA}\end{array}$

Conclusion:

Thus, the current flowing through an element when the charge flow of $q\left(t\right)=5t^2\left(3t^3+4\right)\text{pC}$ is $\underset{\_}{\left(75t^4+40t\right)\text{pA}}$.

To determine

Calculate the current flowing through an element when the charge flow is $q\left(t\right)=2e^{-3t}\sin \left(20\pi t\right)\mu \text{C}$.

Answer to Problem 2P

The current flowing through an element when the charge flow of $q\left(t\right)=2e^{-3t}\sin \left(20\pi t\right)\mu \text{C}$ is $\underset{\_}{\left\{-6e^{-3t}\sin \left(20\pi t\right)+40\pi e^{-3t}\cos \left(20\pi t\right)\right\}\mu \text{A}}$.

Explanation of Solution

Given data:

The charge flow $\left[q\left(t\right)\right]$ is $2e^{-3t}\sin \left(20\pi t\right)\mu \text{C}$.

Calculation:

Substitute $2e^{-3t}\sin \left(20\pi t\right)\mu \text{C}$ for $q\left(t\right)$ in equation (1) to obtain the current flowing through the element as follows:

$\begin{array}{c}i\left(t\right)=\frac{d\left[2e^{-3t}\sin \left(20\pi t\right)\mu \text{C}\right]}{dt}\\ =\left(2\right)\left\{e^{-3t}\frac{d}{dt}\left[\sin \left(20\pi t\right)\right]+\sin \left(20\pi t\right)\frac{d}{dt}\left(e^{-3t}\right)\right\}\mu \text{A}\left\{\because \frac{d}{dt}\left(uv\right)=u\frac{d}{dt}\left(v\right)+v\frac{d}{dt}\left(u\right)\right\}\\ =\left(2\right)\left\{e^{-3t}\left(20\pi \right)\cos \left(20\pi t\right)+\sin \left(20\pi t\right)\left(-3e^{-3t}\right)\right\}\mu \text{A}\left\{\because \frac{d}{dt}\left(\sin at\right)=a\cos t\left(at\right)\right\}\\ =\left[40\pi e^{-3t}\cos \left(20\pi t\right)-6e^{-3t}\sin \left(20\pi t\right)\right]\mu \text{A}\end{array}$$i\left(t\right)=\left[-6e^{-3t}\sin \left(20\pi t\right)+40\pi e^{-3t}\cos \left(20\pi t\right)\right]\mu \text{A}$

Conclusion:

Thus, the current flowing through an element when the charge flow of $q\left(t\right)=2e^{-3t}\sin \left(20\pi t\right)\mu \text{C}$ is $\underset{\_}{\left\{-6e^{-3t}\sin \left(20\pi t\right)+40\pi e^{-3t}\cos \left(20\pi t\right)\right\}\mu \text{A}}$.