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Schaum's Outline of Electric Circuits

Joseph Edminister

Chapter 7

SINUSOIDAL STEADY STATE IN THE FREQUENCY DOMAIN - all with Video Answers

Educators

Chapter Questions

05:05

Problem 1

The current in a series circuit of $R=5 \Omega$ and $L=30 \mathrm{mH}$ lags the applied voltage by $80^{\circ}$. Determine the source frequency and the impedance $\mathbf{Z}$.
From the impedance diagram, Fig. 7-8,
$$
5+j X_L=Z / 800^{\circ} \quad X_L=5 \tan 80^{\circ}=28.4 \Omega
$$

Then $28.4=\omega\left(30 \times 10^{-3}\right)$, whence $\omega=945.2 \mathrm{rad} / \mathrm{s}$ and $f=150.4 \mathrm{~Hz}$.
$$
\mathbf{Z}=5+j 28.4 \Omega
$$

NT
Nikhil Tiwari
Numerade Educator
05:05

Problem 2

At what frequency will the current lead the voltage by $30^{\circ}$ in a series circuit with $R=8 \Omega$ and $C=30 \mu \mathrm{F}$ ?
From the impedance diagram, Fig. 7-9,
$$
8-j X_C=Z L-30^{\circ} \quad-X_C=8 \tan \left(-30^{\circ}\right)=-4.62 \Omega
$$

Then
$$
4.62=\frac{1}{2 \pi f\left(30 \times 10^{-6}\right)} \quad \text { or } \quad f=1149 \mathrm{~Hz}
$$

NT
Nikhil Tiwari
Numerade Educator
09:44

Problem 3

A series $R C$ circuit, with $R=10 \Omega$, has an impedance with an angle of $-45^{\circ}$ at $f_1=500 \mathrm{~Hz}$. Find the frequency for which the magnitude of the impedance is $(a)$ twice that at $f_1$, (b) one-half that at $f_1$.
From $10-j X_C=Z_1 L-45^{\circ}, X_C=10 \Omega$ and $Z_1=14.14 \Omega$.
(a) For twice the magnitude,
$$
10-j X_C=28.28 / \theta_2 \quad \text { or } \quad X_C=\sqrt{(28.28)^2-(10)^2}=26.45 \Omega
$$

Then, since $X_C$ is inversely proportional to $f$,
$$
\frac{10}{26.45}=\frac{f_2}{500} \quad \text { or } \quad f_2=189 \mathrm{~Hz}
$$
(b) A magnitude $Z_3=7.07 \Omega$ is impossible; the smallest magnitude possible is $Z=R=10 \Omega$.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:06

Problem 4

A two-element series circuit has voltage $\mathbf{V}=240 \not 0^{\circ} \quad \mathrm{V}$ and current $\mathbf{I}=50 \angle-60^{\circ} \quad$ A. Determine the current which results when the resistance is reduced to $(a) 30 \%,(b) 60 \%$, of its former value.
(a)
$$
\begin{gathered}
30 \% \times 2.40=0.72 \quad \mathbf{Z}_1=0.72+j 4.16=4.22 \not 80.2^{\circ} \Omega \\
\mathbf{I}_1=\frac{240 \not 0^{\circ}}{4.22 \not 80.2^{\circ}}=56.8 L-80.2^{\circ} \quad \mathbf{A}
\end{gathered}
$$
(b)
$$
\begin{gathered}
60 \% \times 2.40=1.44 \quad \mathbf{Z}_2=1.44+j 4.16=4.40 \not 70.9^{\circ} \Omega \\
\mathbf{I}_2=\frac{240 \not 0^{\circ}}{4.40 \underline{70.9^{\circ}}}=54.5 L-70.9^{\circ} \quad A
\end{gathered}
$$

Ma Ednelyn Lim
Ma Ednelyn Lim
Numerade Educator
03:47

Problem 5

For the circuit shown in Fig. 7-10, obtain $\mathbf{Z}_{\mathrm{eq}}$ and compute $\mathbf{I}$.
For series impedances,
$$
\mathbf{Z}_{\mathrm{eq}}=10 \underline{0^{\circ}}+4.47 / \underline{63.4^{\circ}}=12.0+j 4.0=12.65 \not 18.43^{\circ} \Omega
$$

Then
$$
\mathbf{I}=\frac{\mathbf{V}}{\mathbf{Z}_{\mathrm{eq}}}=\frac{100 / \varphi^{\circ}}{12.65 / 18.43^{\circ}}=7.91 \angle-18.43^{\circ} \quad \mathrm{A}
$$

Prachita Kush
Prachita Kush
Numerade Educator
02:26

Problem 6

Evaluate the impedance $Z_1$ in the circuit of Fig. 7-11.
$$
\mathbf{Z}=\frac{\mathbf{V}}{\mathbf{I}}=20 / 60^{\circ}=10.0+j 17.3 \Omega
$$

Then, since impedances in series add,
$$
5.0+j 8.0+\mathbf{Z}_{\mathrm{t}}=10.0+j 17.3 \quad \text { or } \quad \mathbf{Z}_1=5.0+j 9.3 \Omega
$$

Prachita Kush
Prachita Kush
Numerade Educator
02:26

Problem 7

Compute the equivalent impedance $\mathbf{Z}_{\mathrm{eq}}$ and admittance $\mathbf{Y}_{\mathrm{eq}}$ for the four-branch circuit of Fig. $7-12$.
Using admittances,
$$
\begin{array}{ll}
\mathbf{Y}_1=\frac{1}{j 5}=-j 0.20 \mathrm{~S} & \mathbf{Y}_3=\frac{1}{15}=0.067 \mathrm{~S} \\
\mathbf{Y}_2=\frac{1}{5+j 8.66}=0.05-j 0.087 \mathrm{~S} & \mathbf{Y}_4=\frac{1}{-j 10}=j 0.10 \mathrm{~S}
\end{array}
$$

Then
$$
\mathbf{Y}_{\mathrm{eq}}=\mathbf{Y}_1+\mathbf{Y}_2+\mathbf{Y}_3+\mathbf{Y}_4=0.117-j 0.187=0.221 \_-58.0^{\circ} \mathrm{S}
$$
and
$$
\mathbf{Z}_{\mathrm{eq}}=\frac{1}{\mathbf{Y}_{\mathrm{eq}}}=4.53 / 58.0^{\circ} \Omega
$$

Fig. 7-12

Prachita Kush
Prachita Kush
Numerade Educator

Problem 8

The total current I entering the circuit shown in Fig. 7-12 is 33.0 $-13.0^{\circ}$ A. Obtain the branch current $\mathbf{I}_3$ and the voltage $\mathbf{V}$.
$$
\begin{aligned}
& \mathbf{V}=\mathbf{I} \mathbf{Z}_{\mathrm{eq}}=\left(33.0 \angle-13.0^{\circ}\right)\left(4.53 / 58,0^{\circ}\right)=149.5 / 45.0^{\circ} \mathrm{V} \\
& \mathbf{I}_3=\mathbf{V} \mathbf{Y}_3=\left(149.5 / 45.0^{\circ}\right)\left(\frac{1}{15}\left\langle 0^{\circ}\right)=9.97 \angle 45.0^{\circ} \quad \mathrm{A}\right.
\end{aligned}
$$

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Problem 9

Obtain $\mathbf{Z}_{\mathrm{eq}}$ and $\mathbf{Y}_{\mathrm{eq}}$ for the circuit of Fig. 7-13.
$$
\begin{gathered}
\mathbf{Z}_1=10+j 20=22.4 / 63.43^{\circ} \Omega \quad \mathbf{Z}_2=15-j 15=21.2 L-45^{\circ} \Omega \\
\mathbf{Z}_{\text {eq }}=\frac{\mathbf{Z}_1 \mathbf{Z}_2}{\mathbf{Z}_{\mathrm{t}}+\mathbf{Z}_2}=\frac{\left(22.4 / 63.43^{\circ}\right)\left(21.2 \downharpoonright-45.0^{\circ}\right)}{(10+j 20)+(15-j 15)}=18.63 / 7.12^{\circ} \Omega \\
\mathbf{Y}_{\text {eq }}=\frac{1}{\mathbf{Z}}=0.0537 L-7.12^{\circ} \mathrm{S}
\end{gathered}
$$

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Problem 10

Find $\mathbf{Z}_1$ in the three-branch network of Fig. 7-14, if $\mathbf{I}=31.5 / 24.0^{\circ}$ A for an applied voltage $\mathbf{V}=50.0 / \underline{60.0^{\circ}} \mathrm{V}$.
$$
\mathbf{Y}=\frac{\mathbf{I}}{\mathbf{V}}=0.630-\underline{-36.00^{\circ}}=0.510-j 0.370 \mathrm{~S}
$$

Then
$$
0.510-j 0.370=\mathbf{Y}_1+\frac{1}{10}+\frac{1}{4.0+j 3.0}
$$
whence $\mathbf{Y}_1=0.354 /-45^{\circ} \quad$ S and $\mathbf{Z}_1=2.0+j 2.0 \Omega$.

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06:38

Problem 11

The constants $R$ and $L$ of a coil can be obtained by connecting the coil in series with a known resistance and measuring the coil voltage $V_x$, the resistor voltage $V_1$, and the total voltage $V_T$ (Fig. 7-15). The frequency must also be known, but the phase angles of the voltages are not known. Given that $f=60 \mathrm{~Hz}, V_1=20 \mathrm{~V}, V_x=22.4 \mathrm{~V}$, and $V_T=36.0 \mathrm{~V}$, find $R$ and $L$.
Fig. 7-15
Fig. 7-16
The measured voltages are effective values (see Appendix A); but, as far as impedance calculations are concerned, it makes no difference whether effective or peak values are used.
The (effective) current is $I=V_t / 10=2,0 \mathrm{~A}$. Then
$$
Z_x=\frac{22.4}{2.0}=11.2 \Omega \quad Z_{\text {eq }}=\frac{36,0}{2.0}=18.0 \Omega
$$

From the impedance diagram, Fig. 7-16,
$$
\begin{aligned}
& (18.0)^2=(10+R)^2+(\omega L)^2 \\
& (11.2)^2=R^2+(\omega L)^2
\end{aligned}
$$
where $\omega=2 \pi 60=377 \mathrm{rad} / \mathrm{s}$. Solving simultaneously,
$$
R=4.92 \Omega \quad L=26.7 \mathrm{mH}
$$

Km Neeraj
Km Neeraj
Numerade Educator
02:48

Problem 12

In the parallel circuit shown in Fig. 7-17, the effective values of the currents are: $I_x=18.0 \mathrm{~A}$, $I_1=15.0 \mathrm{~A}, I_T=30.0 \mathrm{~A}$. Determine $R$ and $X_L$.
The problem can be solved in a manner similar to that used in Problem 7.11, but with the admittance diagram.
The (effective) voltage is $V=I_1(4.0)=60.0 \mathrm{~V}$. Then
$$
Y_x=\frac{I_x}{V}=0.300 \mathrm{~S} \quad Y_{\text {eq }}=\frac{I_T}{V}=0.500 \mathrm{~S} \quad Y_{\mathrm{t}}=\frac{1}{4.0}=0.250 \mathrm{~S}
$$
From the admittance diagram, Fig. 7-18,
$$
\begin{aligned}
& (0.500)^2=(0.250+G)^2+B_L^2 \\
& (0.300)^2=G^2+B_L^2
\end{aligned}
$$
which yield $G=0.195 \mathrm{~S}, B_L=0.228 \mathrm{~S}$. Then
$$
R=\frac{1}{G}=5.13 \Omega \quad \text { and } \quad j X_L=\frac{1}{-j B_L}=j 4.39 \Omega
$$
i.e. $X_L=4.39 \Omega$.

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
00:19

Problem 13

Evaluate $V_3$ in Fig. 7-19.
Since the impedances are in series,
$$
\mathbf{V}_3=\left(\frac{\mathbf{Z}_3}{\mathbf{Z}_1+\mathbf{Z}_2+\mathbf{Z}_3}\right) \mathbf{V}_T=\left(\frac{j 15}{16+j 7}\right) 150 / 45^{\circ}=128.81114^{\circ} \quad \mathrm{V}
$$

Amy Jiang
Amy Jiang
Numerade Educator
02:12

Problem 14

Find $\mathbf{I}_1$ and $\mathbf{I}_2$ in the parallel circuit of Fig. 7-20.
In this case, $\mathbf{Z}_{\text {eq }}=\mathbf{Z}_1 \mathbf{Z}_2 /\left(\mathbf{Z}_1+\mathbf{Z}_2\right)$, and the current-division formula becomes
$$
\begin{aligned}
& \mathbf{I}_{\mathbf{t}}=\left(\frac{\mathbf{Z}_2}{\mathbf{Z}_1+\mathbf{Z}_2}\right) \mathbf{I}_T=\left(\frac{10}{13-j 4}\right) 25 / 00^{\circ}=18.4 / 107.1^{\circ} \quad A \\
& \mathbf{I}_2=\left(\frac{\mathbf{Z}}{\mathbf{Z}_1+\mathbf{Z}_2}\right) \mathbf{I}_T=\left(\frac{3-j 4}{13-j 4}\right) 25 / 900^{\circ}=9.19 / 54,0^{\circ} \quad A
\end{aligned}
$$
(or, better, $\mathbf{I}_2=\mathbf{I}_T-\mathbf{I}_1$ ).

Amit Srivastava
Amit Srivastava
Numerade Educator
03:46

Problem 15

Find the phasor voltage $\mathrm{V}_{A B}$ (the potential at point $A$ minus the potential at point $B$ ) in the circuit shown in Fig. 7-21.
The two loops contain currents $\mathbf{I}_1$ and $\mathbf{I}_2$ as shown; hence, no current passes through the $j 20$ reactance.
$$
\mathbf{I}_1=\frac{20 \not 30^{\circ}}{10+j 10}=1.41 \not-15.0^{\circ} \quad \mathrm{A} \quad \mathbf{I}_2=\frac{50 L-45^{\circ}}{5-j 5}=7.07 \not 0^{\circ} \quad \mathrm{A}
$$
Now, $\mathbf{V}_{A B}=\mathbf{V}_{A X}+\mathbf{V}_{X Y}+\mathbf{V}_{Y B}$. In the first loop, $\mathbf{V}_{A X}=\mathbf{V}_S=\mathbf{I}_1(5)=7.07 \angle-15,00^{\circ} \quad \mathrm{V}$. In the connecting branch, $\mathbf{V}_{X Y}=\mathbf{I}(j 20)=0$. And in the second loop, $\mathbf{V}_{Y B}=-\mathbf{V}_s=-\mathbf{I}_2(5)=-35.4 \underline{0}^{\circ} \quad \mathrm{V}$. Then,
$$
\mathrm{V}_{A B}=7.07 \angle-15^{\circ}-35.4 \angle 0^{\circ}=-28.6-j 1.83=28.7 \angle 183.7^{\circ} \quad \mathrm{V}
$$

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
02:51

Problem 16

Obtain the phasor voltage $\mathbf{V}_{A B}$ in the two-branch parallel circuit of Fig. 7-22.
By current-division methods, $\mathbf{I}_1=4.64 \angle 20.1^{\circ} \mathrm{A}$ and $\mathbf{I}_2=17.4 / 30.1^{\circ}$ A. Either path $A X B$ or path $A Y B$ may be considered. Choosing the former,
$$
\mathbf{V}_{A B}=\mathbf{V}_{A X}+\mathbf{V}_{X B}=\mathbf{I}_1(20)-\mathbf{I}_2(j 6)=92.8 \angle 120.1^{\circ}+104.4 \angle L-59.9^{\circ}=11.6 L-59.9^a \quad V
$$

Fig. 7-22
Fig. 7-23

Thane Stiles
Thane Stiles
Numerade Educator
02:47

Problem 17

In the parallel circuit shown in Fig. 7-23, $\mathbf{V}_{A B}=48.3 / 30^{\circ} \quad$ V. Find the applied voltage $\mathbf{V}$.
By voltage division in the two branches:
$$
\mathbf{v}_{A X}=\frac{-j 4}{4-j 4} \mathbf{V}=\frac{1}{1+j} \mathbf{V} \quad \mathbf{v}_{B X}=\frac{j 8.66}{5+j 8.66} \mathbf{V}
$$
and so
$$
\mathbf{V}_{A B}=\mathbf{V}_{A X}-\mathbf{V}_{B X}=\left(\frac{1}{1+j}-\frac{j 8.66}{5+j 8.66}\right) \mathbf{V}=\frac{1}{-0.268+j 1} \mathbf{V}
$$
or
$$
\mathbf{V}=(-0.268+j 1) \mathbf{V}_{A B}=\left(1.035 \angle 05^{\circ}\right)\left(48.3 / 30^{\circ}\right)=50.0 \angle 135^{\circ} \quad \mathrm{V}
$$

Tanishq Gupta
Tanishq Gupta
Numerade Educator
01:54

Problem 18

The voltage-current phasor diagram shown in Fig. 7-24 is for a two-branch parallel circuit. Find the branch impedances and the equivalent impedance.
$$
\begin{aligned}
& \mathbf{Z}_1=\frac{\mathbf{V}}{\mathbf{I}_1}=5.39 / 21.8^{\circ}=5.0+j 2.0 \Omega \\
& \mathbf{Z}_2=\frac{\mathbf{V}}{\mathbf{I}_2}=4.24 L-45.0^{\circ}=3.0-j 3.0 \Omega \\
& \mathbf{Z}_{\mathrm{eq}}=\frac{\mathbf{Z}_1 \mathbf{Z}_2}{\mathbf{Z}_1+\mathbf{Z}_2}=\frac{22.85 L-23.2^{\circ}}{8.0-j 1.0}=2.83-16.2^{\circ} \Omega
\end{aligned}
$$
or $\mathbf{Z}_{\text {eq }}=\mathbf{V} /\left(\mathbf{I}_1+\mathbf{I}_2\right)$.

Carson Merrill
Carson Merrill
Numerade Educator
07:04

Problem 19

Obtain the elements of the series circuit corresponding to the voltage-current phasor diagram Fig. $7-25$, if the frequency is $21.2 \mathrm{kHz}$.
Here the current leads the voltage by $14^{\circ}$ :
$$
\mathbf{Z}_1=\frac{85 /-155^{\circ}}{41.2 L-141^{\circ}}=2,06-14^{\circ}=2.0-j 0.50 \Omega
$$

The elements are a resistance, $R=2.0 \Omega$, and a capacitance such that
$$
X_C=0.50=\frac{1}{2 \pi\left(21.2 \times 10^3\right) C} \quad \text { or } \quad C=15.0 \mu \mathrm{F}
$$

Donald Albin
Donald Albin
Numerade Educator
12:21

Problem 20

Obtain the phasor voltages $\mathbf{V}_{A B}$ and $\mathbf{V}_{B C}$ for the circuit shown in Fig. 7-26.
Using voltage division,
$$
\begin{aligned}
& \mathbf{V}_{A B}=\frac{20 \not-45^{\circ}}{32.4 \not-25.9^{\circ}}\left(100 / \underline{0}^{\circ}\right)=61.8 \not-19.1^{\circ} \quad \mathrm{V} \\
& \mathbf{V}_{B C}=\frac{15 \not 0^{\circ}}{32.4 L-25.9^{\circ}}\left(100 \not 0^{\circ}\right)=46.4 / 25.9^{\circ} \quad \mathrm{V}
\end{aligned}
$$

Linda Winkler
Linda Winkler
Numerade Educator
04:41

Problem 21

A two-element series circuit, with $R=20 \Omega$ and $L=20 \mathrm{mH}$, has an impedance $40.0 \angle \theta \Omega$. Determine the angle $\theta$ and the frequency.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:59

Problem 22

Determine the impedance of the series $R L$ circuit, with $R=25 \Omega$ and $L=10 \mathrm{mH}$, at (a) $100 \mathrm{~Hz},(b)$ $500 \mathrm{~Hz}$, (c) $1000 \mathrm{~Hz}$.

James Kiss
James Kiss
Numerade Educator
02:01

Problem 23

A resistance of $25 \Omega$ is in series with a second circuit element; the circuit frequency is $500 \mathrm{~Hz}$. Find the element if the current ( $a$ ) lags the applied voltage by $20^{\circ} ;(b)$ leads by $20^{\circ}$.

Mayukh Banik
Mayukh Banik
Numerade Educator
02:50

Problem 24

Determine the circuit constants of a two-element series circuit if the applied voltage
$$
v=150 \sin \left(5000 t+45^{\circ}\right) \quad(\mathrm{V})
$$
results in a current $i=3.0 \sin \left(5000 t-15^{\circ}\right)$ (A). Ans. $25 \Omega, 8.66 \mathrm{mH}$

Nick Johnson
Nick Johnson
Numerade Educator
03:39

Problem 25

A series circuit of $R=10 \Omega$ and $C=40 \mu \mathrm{F}$ has an applied voltage $v=500 \cos \left(2500 t-20^{\circ}\right)$ (V). Find the resulting current $i$. Ans. $25 \sqrt{2} \cos \left(2500 t+25^{\circ}\right)$ (A)

Kajal Gautam
Kajal Gautam
Numerade Educator

Problem 26

Figure 7-27 is the voltage-current phasor diagram for a two-element series circuit at angular frequency $300 \mathrm{rad} / \mathrm{s}$. Find the elements if the phasor magnitudes are $200 \mathrm{~V}$ and $20 \mathrm{~A}$. Ans. $8 \Omega, 20 \mathrm{mH}$

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00:52

Problem 27

Three impedances are in series: $\mathbf{Z}_1=3.0 / 45^{\circ} \quad \Omega, \mathbf{Z}_2=10 \sqrt{2} / 45^{\circ} \Omega, \mathbf{Z}_3=5.0 \angle-90^{\circ} \Omega$. Find the applied voltage $\mathbf{V}$, if the voltage across $\mathbf{Z}_1$ is $27.0 \angle-10^{\circ}$ V. Ans. $126.5 /-24.6^{\circ}$ V

Ze-Han Lee
Ze-Han Lee
Numerade Educator
02:10

Problem 28

For the three-element series circuit in Fig. 7-28, (a) find the current $\mathbf{I} ;(b)$ find the voltage across cach impedance and construct the voltage phasor diagram which shows that $V_1+V_2+V_3=100 / 0^{\circ} \quad V$.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator

Problem 29

Find $\mathrm{Z}$ in the series circuit shown in Fig. 7-30, if $\mathrm{V}=13.05 / 15.0^{\circ} \quad \mathrm{V} . \quad$ Ans. $4.0-j 15.0 \quad \Omega$

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02:36

Problem 30

Determine the equivalent impedance and admittance of the four-branch parallel circuit shown in Fig. 7-31. Ans. $4.54 / 57.9^{\circ} \quad \Omega, 0.220 /-57.9^{\circ}$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
03:27

Problem 31

Find $\mathbf{Z}$ in the parallel circuit of Fig. 7-32, if $\mathbf{V}=50,0 / 30,0^{\circ} \quad \mathrm{V}$ and $\mathbf{I}=27.9 / 57.8^{\circ} \mathrm{A}$. Ans. $5.0 /-30^{\circ} \Omega$
Fig. 7-32

Sheh Lit Chang
Sheh Lit Chang
University of Washington
03:35

Problem 32

Obtain the conductance and susceptance corresponding to a voltage $\mathrm{V}=85.0 / 205^{\circ} \quad \mathrm{V}$ and a resulting current $\mathbf{I}=41.2 /-141.0^{\circ}$ A. Ans. $0.471 \mathrm{~S}, 0.117 \mathrm{~S}$ (capacitive)

Vishal Gupta
Vishal Gupta
Numerade Educator
02:23

Problem 33

A two-branch parallel circuit, with branch impedances
$$
\mathbf{Z}_1=15.0 / 20.0^{\circ} \Omega \quad \mathbf{Z}_2=20.0 / 45.0^{\circ} \Omega
$$
has an applied voltage $\mathrm{V}=240 / 45.00^{\circ} \mathrm{V}$. Determine the total current by first obtaining the equivalent admittance.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
02:34

Problem 34

An admittance consists of $1.77 \mathrm{mS}$ conductance and $1.77 \mathrm{mS}$ inductive susceptance. Find the resistance and reactance of the corresponding impedance. Ans. $282 \Omega, 282 \Omega$ (inductive)

Aja S
Aja S
Numerade Educator
01:59

Problem 35

For the series-parallel circuit shown in Fig. 7-33, obtain the equivalent admittance.

Amit Srivastava
Amit Srivastava
Numerade Educator
03:21

Problem 36

A practical coil contains resistance as well as inductance and can be represented by either a series or parallel circuit, as suggested in Fig. 7-34. Obtain $R_p$ and $L_p$ in terms of $R_s$ and $L_s$.

Mahnoor Khan
Mahnoor Khan
Numerade Educator
01:55

Problem 37

Obtain $\mathrm{V}_1$ in Fig. 7-35, given an applied voltage V=150/-45 V.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
06:18

Problem 38

Five impedances are scries connected. Impedance $\mathbf{Z}_1=15,0 / 30^{\circ} \quad \Omega$ has a voltage $\mathbf{V}_1=2.5 /-15^{\circ} \quad V$. Find $\mathbf{V}_4$ for $\mathbf{Z}_4=6.5 / 4 \varphi^{\circ} \Omega$. Ans. $1.08 /-45^{\circ} \quad \mathrm{V}$

NT
Nikhil Tiwari
Numerade Educator
01:47

Problem 39

Obtain the currents $\mathbf{I}_1$ and $\mathbf{I}_2$ in Fig. 7-36. Ans. $3.80 /-64.3^{\circ}$ A, $2.85 /-89.3^{\circ} \quad$ A
Fig. 7-36
Fig. 7-37

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
01:16

Problem 40

Referring to Fig. 7-13, obtain the total current if the current in the $15-j 15$ branch is $2.5 /-90^{\circ} \mathrm{mA}$. Ans. $2.85-142.1^{\circ} \mathrm{mA}$

Prabhu Ramji
Prabhu Ramji
Numerade Educator

Problem 41

Three parallel admittances are shown in Fig. 7.37. Obtain the three branch currents. Ans. $2.24 / 31.16^{\circ}$ A, $7.46 / 61.16^{\circ}$ A, $12.2 \angle-38.84^{\circ}$ A

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10:20

Problem 42

A two-branch parallel circuit, with $\mathbf{Z}_1=9 / 90^{\circ} \Omega$, divides $\mathbf{I}_T=5.0 / 45^{\circ}$ A such that $\mathbf{I}_2=3.0 / 98.13^{\circ} \quad$ A. Find $\mathbf{Z}_2$. Ans. $12.0+j 0 \Omega$

Abid Hussain
Abid Hussain
Numerade Educator
08:56

Problem 43

In the network shown in Fig. 7-38, find the voltage $\mathrm{V}_{A B .}$ Ans. $5.96 / 105^{\circ} \mathrm{V}$
$3 \Omega$
$3 \Omega$

Susan Hallstrom
Susan Hallstrom
Numerade Educator
02:45

Problem 44

A series combination of $R$ and $C$ is in parallel with a resistance of $20.0 \Omega$. At a source frequency of $60 \mathrm{~Hz}$, the total current of 7.02 A (amplitude) divides so that the $20.0 \Omega$ resistor takes $6.0 \mathrm{~A}$ and the $R C$ branch $2.3 \mathrm{~A}$ (amplitudes). Evaluate $R$ and $C \quad$ Ans. $15.0 \Omega, 53.2 \mu \mathrm{F}$

Zulfiqar Ali
Zulfiqar Ali
Numerade Educator
01:26

Problem 45

In the network shown in Fig. $7-39$ the $60-\mathrm{Hz}$ current magnitudes are known to be: $I_T=29.9 \mathrm{~A}$, $I_1=22.3 \mathrm{~A}$, and $I_2=8.0 \mathrm{~A}$. Obtain the circuit constants $R$ and $L . \quad$ Ans. $5.8 \Omega, 38.5 \mathrm{mH}$

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
00:52

Problem 46

Obtain the magnitude of the voltage $V_{A B}$ in the two-branch parallel network of Fig. $7-40$, if $X_L$ is (a) $5 \Omega$, (b) $15 \Omega$, (c) $0 \Omega$. Ans. $50 \mathrm{~V}$, whatever $X_L$.

Ze-Han Lee
Ze-Han Lee
Numerade Educator

Problem 47

In the network shown in Fig. 7-41, $V_{A B}=36.1 / 3.18^{\circ} \quad V$. Find the source voltage $V$.

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