Chapter 7, Problem 1E

Making use of the passive sign convention, determine the current flowing through a 100 pF capacitor for t ≥ 0 if its voltage v_C(t) is given by (a) 5 V; (b) 10e^−t V; (c) 2 sin 0.01t V; (d) −5 + 2 sin 0.01t V.

To determine

Find the current flowing through a $100\text{ pF}$ capacitor for $t\ge 0\text{ s}$.

Answer to Problem 1E

The current flowing through the capacitor is $0\text{ A}$.

Explanation of Solution

Given Data:

The voltage across the capacitor is $5\text{ V}$.

The capacitance of the capacitor is $100\text{ pF}$.

Formula used:

The expression for the current in the capacitor is given as follows,

$i_C=\frac{Cd{v}_C\left(t\right)}{dt}$ (1)

Here,

$i_C$ is the current passing through the capacitor.

$C$ is the capacitance of the capacitor.

$v_C\left(t\right)$ is the voltage across the capacitor.

Calculation:

Substitute $100\text{ pF}$ for $C$ and $5\text{ V}$ for $v_C\left(t\right)$ in equation (1).

$\begin{array}{c}{i}_C=\left(100\text{ pF}\right)\frac{d\left(5\text{ V}\right)}{dt}\\ =0\text{ A}\end{array}$

Conclusion:

Thus, the current flowing through the capacitor is $0\text{ A}$.

To determine

The current flowing through a $100\text{ pF}$ capacitor for $t\ge 0\text{ s}$.

Answer to Problem 1E

The current flowing through the capacitor is $\left(-{10}^{-9}\right)e^{-t}\text{ A}$.

Explanation of Solution

Given Data:

The voltage across the capacitor is $10e^{-t}\text{ V}$.

The capacitance of the capacitor is $100\text{ pF}$.

Calculation:

Substitute $100\text{ pF}$ for $C$ and $10e^{-t}\text{ V}$ for $v_C\left(t\right)$ in equation (1).

$\begin{array}{c}{i}_C=\left(100\text{ pF}\right)\frac{d\left(10e^{-t}\text{ V}\right)}{dt}\\ =\left(100\times {10}^{-12}\text{ F}\right)\left(-10e^{-t}\frac{\text{V}}{s}\right)\left\{\because 1\text{ pF}=\text{1}\times {\text{10}}^{-12}\text{ F}\right\}\\ =\left(-{10}^{-9}\right)e^{-t}\text{ A}\end{array}$

Conclusion:

Thus, the current flowing through the capacitor is $\left(-{10}^{-9}\right)e^{-t}\text{ A}$.

To determine

The current flowing through a $100\text{ pF}$ capacitor for $t\ge 0\text{ s}$.

Answer to Problem 1E

The current flowing through the capacitor is $2\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ pA}$.

Explanation of Solution

Given Data:

The voltage across the capacitor is $2\left(\sin \left(\text{0}\text{.01}t\right)\right)\text{ V}$.

The capacitance of the capacitor is $100\text{ pF}$.

Calculation:

Substitute $100\text{ pF}$ for $C$ and $2\left(\sin \left(\text{0}\text{.01}t\right)\right)\text{ V}$ for $v_C\left(t\right)$ in equation (1).

$\begin{array}{c}{i}_C=\left(100\text{ pF}\right)\frac{d\left(2\left(\sin \left(\text{0}\text{.01}t\right)\right)\text{ V}\right)}{dt}\\ =\left(100\times {10}^{-12}\text{ F}\right)\left(2\times 0.01\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\frac{\text{V}}{s}\right)\left\{\because 1\text{ pF}=\text{1}\times {\text{10}}^{-12}\text{ F}\right\}\\ =2\times {10}^{-12}\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ A}\\ =2\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ pA }\left\{\because 1\text{ A}=\text{1}\times {\text{10}}^{12}\text{ pA}\right\}\end{array}$

Conclusion:

Thus, the current flowing through the capacitor is $2\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ pA}$.

To determine

The current flowing through a $100\text{ pF}$ capacitor for $t\ge 0\text{ s}$.

Answer to Problem 1E

The current flowing through the capacitor is $2\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ pA}$.

Explanation of Solution

Given Data:

The voltage across the capacitor is $-5+2\left(\sin \left(0.01t\right)\right)\text{ V}$.

The capacitance of the capacitor is $100\text{ pF}$.

Calculation:

Substitute $100\text{ pF}$ for $C$ and $-5+2\left(\sin \left(0.01t\right)\right)\text{ V}$ for $v_C\left(t\right)$ in equation (1).

$\begin{array}{c}{i}_C=\left(100\text{ pF}\right)\frac{d\left(-5+2\left(\sin \left(0.01t\right)\right)\text{ V}\right)}{dt}\\ =\left(100\times {10}^{-12}\text{ F}\right)\left(2\times 0.01\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\frac{\text{V}}{s}\right)\left\{\because 1\text{ pF}=\text{1}\times {\text{10}}^{-12}\text{ F}\right\}\\ =2\times {10}^{-12}\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ A}\\ =2\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ pA }\left\{\because 1\text{ A}=\text{1}\times {\text{10}}^{12}\text{ pA}\right\}\end{array}$

Conclusion:

Thus, the current flowing through the capacitor is $2\left(\text{cos }\left(\text{0}\text{.01}t\right)\right)\text{ pA}$.