Chapter 11, Problem 1E

(a)

To determine

Find the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$, when $v_s=9\text{V}$.

Answer to Problem 1E

The instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ is 3.24 W.

Explanation of Solution

Calculation:

Refer to the figure given in the question.

Consider the expression of instantaneous power delivered to the resistor.

$p_{1\Omega }=\frac{v_{1\Omega }^2}{1\Omega }$        (1)

Here,

$v_{1\Omega }$ is the voltage across $1\Omega$ resistor.

Find the voltage across $1\Omega$ resistor using voltage divider rule.

$v_{1\Omega }=\frac{v_s}{4+1}\left(1\right)$

$v_{1\Omega }=\frac{v_s}{5}$        (2)

Substitute 9 V for $v_s$ in equation (2).

$v_{1\Omega }=\frac{9}{5}\text{V}$

Substitute $\frac{9}{5}\text{V}$ for $v_{1\Omega }$ in equation (1).

$\begin{array}{c}{p}_{1\Omega }=\frac{{\left(\frac{9}{5}\text{V}\right)}^2}{1\Omega }\\ =\frac{81}{25}\text{W}\\ =\frac{81}{25}\text{W}\\ =3.24\text{W}\end{array}$

Since the power value is independent to the time. Hence, the instantaneous power delivered to the resistor is same (3.24 W) for all values of time.

Conclusion:

Thus, the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ is 3.24 W.

(b)

To determine

Find the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$, when $v_s=9\sin 2t\text{V}$.

Answer to Problem 1E

The instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ are,

t in s	$p_{1\Omega }\left(t\right)=\frac{81{\left(\sin 2t\right)}^2}{25}\text{W}$
0	$p_{1\Omega }\left(0\text{s}\right)=\text{0}\text{W}$
1	$p_{1\Omega }\left(1\text{s}\right)=2.68\text{W}$
2	$p_{1\Omega }\left(2\text{s}\right)=1.856\text{W}$

Explanation of Solution

Calculation:

Substitute $9\sin 2t\text{V}$ for $v_s$ in equation (2).

$v_{1\Omega }=\frac{9\sin 2t}{5}\text{V}$

Substitute $\frac{9\sin 2t}{5}\text{V}$ for $v_{1\Omega }$ in equation (1).

$\begin{array}{c}{p}_{1\Omega }\left(t\right)=\frac{{\left(\frac{9\sin 2t}{5}\text{V}\right)}^2}{1\Omega }\\ =\frac{81{\left(\sin 2t\right)}^2}{25}\text{W}\end{array}$

The power values for various values of time are tabulated below.

Table 1

t in s	$p_{1\Omega }\left(t\right)=\frac{81{\left(\sin 2t\right)}^2}{25}\text{W}$
0	$\begin{array}{c}{p}_{1\Omega }\left(0\text{s}\right)=\frac{81{\left(\sin 2\left(0\right)\right)}^2}{25}\text{W}\\ =\text{0}\text{W}\end{array}$
1	$\begin{array}{c}{p}_{1\Omega }\left(1\text{s}\right)=\frac{81{\left(\sin 2\left(1\right)\right)}^2}{25}\text{W}\\ =2.68\text{W}\end{array}$
2	$\begin{array}{c}{p}_{1\Omega }\left(2\text{s}\right)=\frac{81{\left(\sin 2\left(2\right)\right)}^2}{25}\text{W}\\ =1.856\text{W}\end{array}$

Conclusion:

Thus, the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ are calculated.

(c)

To determine

Find the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$, when $v_s=9\sin \left(2t+13°\right)\text{V}$.

Answer to Problem 1E

The instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ are,

t in s	$p_{1\Omega }\left(t\right)=\frac{81{\left(\sin \left(2t+\frac{13\pi }{180}\right)\right)}^2}{25}\text{W}$
0	$0.164\text{W}$
1	$2.03\text{W}$
2	$2.53\text{W}$

Explanation of Solution

Calculation:

Substitute $9\sin \left(2t+13°\right)\text{V}$ for $v_s$ in equation (2).

$v_{1\Omega }=\frac{9\sin \left(2t+13°\right)}{5}\text{V}$

Substitute $9\sin \left(2t+13°\right)\text{V}$ for $v_{1\Omega }$ in equation (1).

$\begin{array}{c}{p}_{1\Omega }\left(t\right)=\frac{{\left(\frac{9\sin \left(2t+13°\right)}{5}\text{V}\right)}^2}{1\Omega }\\ =\frac{81{\left(\sin \left(2t+13°\right)\right)}^2}{25}\text{W}\end{array}$

Convert $13°$ into radians for easy calculation.

$\begin{array}{c}{p}_{1\Omega }\left(t\right)=\frac{81{\left(\sin \left(2t+13\times \frac{\pi }{180}\right)\right)}^2}{25}\text{W}\\ =\frac{81{\left(\sin \left(2t+\frac{13\pi }{180}\right)\right)}^2}{25}\text{W}\end{array}$

The power values for various values of time are tabulated below.

Table 2

t in s	$p_{1\Omega }\left(t\right)=\frac{81{\left(\sin \left(2t+\frac{13\pi }{180}\right)\right)}^2}{25}\text{W}$
0	$p_{1\Omega }\left(0\text{s}\right)=0.164\text{W}$
1	$p_{1\Omega }\left(1\text{s}\right)=2.03\text{W}$
2	$p_{1\Omega }\left(2\text{s}\right)=2.53\text{W}$

Conclusion:

Thus, the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ are calculated.

(d)

To determine

Find the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$, when $v_s=9e^{-t}\text{V}$.

Answer to Problem 1E

The instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ are,

t in s	$p_{1\Omega }\left(t\right)=\frac{81e^{-2t}}{25}\text{W}$
0	$3.24\text{W}$
1	$0.44\text{W}$
2	$0.06\text{W}$

Explanation of Solution

Calculation:

Substitute $9e^{-t}\text{V}$ for $v_s$ in equation (2).

$v_{1\Omega }=\frac{9e^{-t}}{5}\text{V}$

Substitute $\frac{9e^{-t}}{5}\text{V}$ for $v_{1\Omega }$ in equation (1).

$\begin{array}{c}{p}_{1\Omega }\left(t\right)=\frac{{\left(\frac{9e^{-t}}{5}\text{V}\right)}^2}{1\Omega }\\ =\frac{81e^{-2t}}{25}\text{W}\end{array}$

The power values for various values of time are tabulated below.

Table 3

t in s	$p_{1\Omega }\left(t\right)=\frac{81e^{-2t}}{25}\text{W}$
0	$p_{1\Omega }\left(0\text{s}\right)=3.24\text{W}$
1	$p_{1\Omega }\left(1\text{s}\right)=0.44\text{W}$
2	$p_{1\Omega }\left(2\text{s}\right)=0.06\text{W}$

Conclusion:

Thus, the instantaneous power delivered to the $1\Omega$ resistor at $t=0,t=1\text{s},\text{and}t=2\text{s}$ are calculated.