A voltage v(t) = 100 cos(60t + 20) V is applied to a parallel combination of a 40-k

Given the sinusoid \(45 \cos \left(5 \pi t+36^{\circ}\right)\), calculate its amplitude, phase, angular frequency, period, and frequency. Equation Transcription: Text Transcription: 45 cos(5 pi t + 36 degree)

Find the phase angle between \(i_{1}=-4 \sin \left(377 t+55^{\circ}\right) \quad \text { and } \quad i_{2}=5 \cos \left(377 t-65^{\circ}\right)\) Does \(i_{1}\) lead or lag \(i_{2}\) ? Equation Transcription: Text Transcription: i_1= ?4 sin(377t + 55) i_2= 5 cos(377t ? 65) i_1 i_2

Evaluate the following complex numbers: (a) \(\left[(5+\mathrm{j} 2)(-1+\mathrm{j} 4)-5 \angle 60^{\circ}\right]^{*}\) (b) \(\frac{10+j 5+3 \angle 40^{\circ}}{-3+j 4}+10 \angle 30^{\circ}+j 5\) Equation Transcription: Text Transcription: [(5 + j2)(-1 + j4)-5 angle 60 degree ]* 10 +j5 + 3 angle 40 degree/-3 + j4+10 angle 30 degree+ j5

Express these sinusoids as phasors: (a) \(v=-14 \sin \left(5 t-22^{\circ}\right) \mathrm{V}\) (b) \(i=-8 \cos \left(16 t+15^{\circ}\right) \mathrm{A}\) Equation Transcription: Text Transcription: v = ?14 sin(5t ? 22 degree) V i = ?8 cos(16t + 15 degree) A

Find the sinusoids corresponding to these phasors: (a) \(\mathrm{V}=-25 \angle 40^{\circ} \mathrm{V}\) (b) \(I=j(12-j 5) A\) Equation Transcription: Text Transcription: V=-25 angle 40 degree V I = j(12-j5) A

If \(v_{1}=-10 \sin \left(\omega t-30^{\circ}\right) \mathrm{V}\) and \(v_{2}=20 \cos \left(\omega t+45^{\circ}\right) \mathrm{V}\), find \(v=v_{1}+v_{2}\). Equation Transcription: Text Transcription: v_1= ?10 sin(watt t ? 30 degree) V v_2= 20 cos(watt t + 40 degree) V v = v_1 +v_2

Find the voltage \(v(t)\) in a circuit described by the inte grodifferential equation \(2 \frac{d v}{d t}+5 v+10 \int v d t=50 \cos \left(5 t-30^{\circ}\right)\) using the phasor approach. Equation Transcription: Text Transcription: v(t) 2dv/dt+5v +10 integral v dt=50 cos(5t-30 degree)

If voltage \(v=25 \sin \left(100 t-15^{\circ}\right) V\) is applied to a \(50 \mu \mathrm{F}\) capacitor, calculate the current through the capacitor. Equation Transcription: Text Transcription: v = 25 sin(100t-15 degree) V 50 mu F

Refer to Fig. 9.17. Determine \(v(t)\) and \(i(t)\). Equation Transcription: Text Transcription: v(t) i(t)

Determine the input impedance of the circuit in Fig. 9.24 at \(\omega=\) \(20 \mathrm{rad} / \mathrm{s}\). Equation Transcription: Text Transcription: watt= 20 rad/s

Calculate \(v_{0}\) in the circuit of Fig. 9.27. Equation Transcription: Text Transcription: v_0

Find \(I\) in the circuit of Fig. 9.30. Equation Transcription: Text Transcription: I

Design an \(R C\) circuit to provide a \(90^{\circ}\) lagging phase shift of the output voltage relative to the input voltage. If an ac voltage of \(60 \mathrm{~V} \mathrm{rms}\) is applied, what is the output voltage? Equation Transcription: 90° Text Transcription: RC 90 degree V

Refer to the \(RL\) circuit in Fig. 9.36. If \(10\ V\) is applied to the input, find the magnitude and the phase shift produced at \(5\ kHz\). Specify whether the phase shift is leading or lagging. Equation Transcription: Text Transcription: RL 10 V 5 kHz

In the ac bridge circuit of Fig. 9.37, suppose that balance is achieved when \(\mathbf{Z}_{1}\) is a \(4.8-\mathrm{k} \Omega\) resistor, \(\mathbf{Z}_{2}\) is a 10- \(\Omega\) resistor in series with a \(0.25-\mu \mathrm{H}\) inductor, \(\mathbf{Z}_{3}\) is a \(12-\mathrm{k} \Omega\) resistor, and \(f=6 \mathrm{MHz}\). Determine the series components that make up \(\mathbf{Z}_{x}\). Equation Transcription: Text Transcription: Z_1 Z_2 Z_3 Z_x 4.8-k ohms 10-k ohms 12-k ohms 0.25-mu H f = 6 MHz

Transform the following sinusoids to phasors: (a) \(-20 \cos \left(4 t+135^{\circ}\right)\) (b) \(8 \sin \left(20 t+30^{\circ}\right)\) (c) \(20 \cos (2 t)+15 \sin (2 t)\) Equation Transcription: Text Transcription: -20 cos(4t + 135 degree) 8 sin(20t + 30 degree) 20 cos (2t) + 15 sin (2t)

Two voltages \(v_{1}\) and \(v_{2}\) appear in series so that their sum is \(v=v_{1}+v_{2}\). If \(v_{1}=10 \cos (50 t-\pi / 3) \mathrm{V}\) and \(v_{2}=12 \cos \left(50 t+30^{\circ}\right) \mathrm{V}\), find \(v\). Equation Transcription: Text Transcription: v_1 v_2 v = v_1 + v_2 v_1 = 10 cos(50t - pi/3) V v_2 = 12 cos(50t + 30 degree) V v

Obtain the sinusoids corresponding to each of the following phasors: (a) \(\mathbf{V}_{1}=60 / 15^{\circ} \mathrm{V}, \omega=1\) (b) \(\mathbf{V}_{2}=6+j 8 \mathrm{~V}, \omega=40\) (c) \(\mathbf{I}_{1}=2.8 e^{-j \pi / 3} \mathrm{~A}, \omega=377\) (d) \(\mathbf{I}_{2}=-0.5-j 1.2 \mathrm{~A}, \omega=10^{3}\) Equation Transcription: Text Transcription: V_1=60 angle 15 degree V, watt=1 V_2=6+j8 V, watt=40 I_1=2.8e^-j pi/3A,watt=377 I_2=-0.5-j1.2 A, watt=10^3

Using phasors, find: (a) \(3 \cos \left(20 t+10^{\circ}\right)-5 \cos \left(20 t-30^{\circ}\right)\) (b) \(40 \sin 50 t+30 \cos \left(50 t-45^{\circ}\right)\) (c) \(20 \sin 400 t+10 \cos \left(400 t+60^{\circ}\right)\) \(-5 \sin \left(400 t-20^{\circ}\right)\) Equation Transcription: Text Transcription: 3 cos(20t + 10 degree)-5 cos(20t-30 degree) 40 sin 50t + 30 cos(50t-45 degree) 20 sin 400t + 10 cos(400t + 60 degree)-5 sin(400t-20 degree)

A linear network has a current input\(7.5 \cos \left(10 t+30^{\circ}\right) \mathrm{A}\) and a voltage output \(120 \cos \left(10 t+75^{\circ}\right) \mathrm{V}\). Determine the associated impedance. Equation Transcription: Text Transcription: 7.5 cos(10t + 30 degree)A 120 cos(10t + 75 degree) V

Simplify the following: (a) \(f(t)=5 \cos \left(2 t+15^{\circ}\right)-4 \sin \left(2 t-30^{\circ}\right)\) (b) \(g(t)=8 \sin t+4 \cos \left(t+50^{\circ}\right)\) (c) \(h(t)=\int_{0}^{t}(10 \cos 40 t+50 \sin 40 t) d t\) Equation Transcription: Text Transcription: f(t)=5 cos(2t+15 degree)-4 sin(2t-30 degree) g(t)=8 sin t+4 cos(t+50 degree) h(t)=integral_0^t(10 cos 40 t+50 sin 40t)dt

An alternating voltage is given by \(v(t)= 55 \cos \left(5 t+45^{\circ}\right) V\). Use phasors to find \(10 v(t)+4 \frac{d v}{d t}-2 \int_{-\infty}^{t} v(t) d t\) Assume that the value of the integral is zero at \(t=-\infty\). Equation Transcription: Text Transcription: v(t) = 55 cos(5t + 45 degree) V 10v(t)+4dv/dt-2 integral_-infinity^t v(t) dt t=-infinity

Apply phasor analysis to evaluate the following: (a) \(v=\left[110 \sin \left(20 t+30^{\circ}\right)+220 \cos \left(20 t-90^{\circ}\right)\right] \mathrm{V}\) (b) \(i=\left[30 \cos \left(5 t+60^{\circ}\right)-20 \sin \left(5 t+60^{\circ}\right)\right] \mathrm{A}\) Equation Transcription: Text Transcription: v = [110 sin(20t + 30 degree) + 220 cos(20t ? 90 degree)] V i = [30 cos(5t + 60 degree) ? 20 sin(5t + 60 degree)] A

Find \(v(t)\) in the following integro differential equations using the phasor approach: (a) \(v(t)+\int v d t=10 \cos t\) (b) \(\frac{d v}{d t}+5 v(t)+4 \int v d t=20 \sin \left(4 t+10^{\circ}\right)\) Equation Transcription: Text Transcription: v(t) v(t)+integral v dt=10 cos t dv/dt+5v(t)+4 integral v dt=20 sin(4t+10 degree)

Using phasors, determine \(i(t)\) in the following equations: (a) \(2 \frac{d i}{d t}+3 i(t)=4 \cos \left(2 t-45^{\circ}\right)\) (b) \(10 \int i d t+\frac{d i}{d t}+6 i(t)=5 \cos \left(5 t+22^{\circ}\right) \mathrm{A}\) Equation Transcription: Text Transcription: i(t) 2 di/dt+3i(t)=4 cos(2t-45 degree) 10 integral i dt+di/dt+6i(t)=5 cos(5t+22 degree)A

The loop equation for a series \(R L C\) circuit gives \(\frac{d i}{d t}+2 i+\int_{-\infty}^{t} i d t=\cos 2 t \mathrm{~A}\) Assuming that the value of the integral at \(t=-\infty\) is zero, find \(i(t)\) using the phasor method. Equation Transcription: Text Transcription: di/dt+2i+integral_-infinity^t i dt=cos 2t A t=-infinity

A parallel \(R L C\) circuit has the node equation \(\frac{d v}{d t}+50 v+100 \int v d t=110 \cos \left(377 t-10^{\circ}\right) \mathrm{V}\) Determine \(v(t)\) using the phasor method. You may assume that the value of the integral at \(t=-\infty\) is zero. Equation Transcription: Text Transcription: dv/dt+50v+100 integral v dt=110 cos(377t-10 degree) V v(t t=-infinity

Determine the current that flows through an \(20- \Omega\) resistor connected to a voltage source \(v_{\mathrm{s}}=120 \cos \left(377 t+37^{\circ}\right) \mathrm{V}\). Equation Transcription: Text Transcription: 20-ohms v_s= 120 cos (377t + 37 degree) V

Given that \(v_{c}(0)=2 \cos \left(155^{\circ}\right) \mathrm{V}\), what is the instantaneous voltage across a \(2-\mu \mathrm{F}\) capacitor when the current through it is \(i=4 \sin \left(10^{6} t+25^{\circ}\right) \mathrm{A}\) ? Equation Transcription: Text Transcription: v_c(0) = 2 cos(155 degree) V 2-mu F i = 4 sin(106t + 25 degree ) A

A voltage \(v(t)=100 \cos \left(60 t+20^{\circ}\right) \mathrm{V}\) is applied to a parallel combination of a \(40-\mathrm{k} \Omega \) resistor and a \(50-\mu \mathrm{F}\) capacitor. Find the steady-state currents through the resistor and the capacitor. Equation Transcription: Text Transcription: v(t) = 100 cos(60t + 20 degree) V 40-k ohms 50-mu F

A series \(R L C\) circuit has \(R=80 \Omega, L=240 \mathrm{mH}\), and \(C=5 \mathrm{mF}\). If the input voltage is \(v(t)=115 \cos 2 t\), find the current flowing through the circuit. Equation Transcription: Text Transcription: RLC R = 80 ohms, L = 240 mH C = 5 mF v(t) = 115 cos 2t

A series \(RL\) circuit is connected to a \(220-V\) ac source. If the voltage across the resistor is \(170\ V\), find the voltage across the inductor. Equation Transcription: Text Transcription: RL 220-V 170 V

A series RL circuit is connected to a 220-V ac source. If the voltage across the resistor is 170 V, find the voltage across the inductor.

What value of \(\omega\) will cause the forced response, \(v_{o}\), in Fig.9.41 to be zero? Equation Transcription: Text Transcription: watt v_o

Find the steady-state current $i$ in the circuit of Fig. 9.42, when \(v_{s}(t)=115 \cos 200 t \mathrm{~V}\). Equation Transcription: Text Transcription: v_s(t) = 115 cos 200t V

Using Fig. 9.43, design a problem to help other students better understand impedance.

Determine the admittance \(\mathbf{Y}\) for the circuit in Fig. 9.44. Equation Transcription: Text Transcription: Y

Using Fig. 9.45, design a problem to help other students better understand admittance.

For the circuit shown in Fig. 9.46, find \(Z_{\mathrm{eq}}\) and use that to find the current \(\mathbf{I}\). Let \(\omega=10 \mathrm{rad} / \mathrm{s}\). Equation Transcription: ???? Text Transcription: Z_eq I watt= 10 rad/s

In the circuit of Fig. 9.47, find \(i_{o}\), when: (a) \(\omega=1 \mathrm{rad} / \mathrm{s}\) (b) \(\omega=5 \mathrm{rad} / \mathrm{s}\) (c) \(\omega=10 \mathrm{rad} / \mathrm{s}\) Equation Transcription: Text Transcription: i_o watt= 1 rad/s watt= 5 rad/s watt= 10 rad/s

Find \(v(t)\) in the \(RLC\) circuit of Fig. 9.48. Equation Transcription: Text Transcription: v(t) RLC

Calculate \(v_{o}(t)\) in the circuit of Fig. 9.49. Equation Transcription: Text Transcription: v_o(t)

Find current \(\mathbf{I}_{o}\) in the circuit shown in Fig.9.50. Equation Transcription: Text Transcription: I_o

Calculate \(i(t)\) in the circuit of Fig. 9.51. Equation Transcription: Text Transcription: i(t)

Find current \(\mathbf{I}_{o}\) in the network of Fig. 9.52. Equation Transcription: Text Transcription: I_o

If \(v_{s}=100 \sin \left(10 t+18^{\circ}\right) \mathrm{V}\) in the circuit of Fig. 9.53, find \(i_{o}\). Equation Transcription: Text Transcription: v_s= 100 sin(10t + 18 degree) V i_o

In the circuit of Fig. 9.54, determine the value of \(i_{s}(t)\). Equation Transcription: Text Transcription: i_s(t)

Given that \(v_{s}(t)=20 \sin \left(100 t-40^{\circ}\right)\) in Fig. 9.55, determine \(i_{x}(t)\). Equation Transcription: Text Transcription: v_s= 20 sin(100t - 40 degree) i_x(t)

Find \(v_{s}(t)\) in the circuit of Fig.9.56 if the current \(i_{x}\) through the \(1-\Omega\) resistor is \(8 \sin 200 t\) A. Equation Transcription: Text Transcription: v_s(t) 1-ohms 8 sin 200t A

Determine \(v_{x}\) in the circuit of Fig. 9.57. Let \(i_{s}(t)= 5 \cos \left(100 t+40^{\circ}\right) \mathrm{A}\). Equation Transcription: Text Transcription: v_x i_s(t) = 5 cos(100t + 40 degree) A

If the voltage \(v_{o}\) across the \(2-\Omega\) resistor in the circuit of Fig. 9.58 is \(90 \cos 2 t \mathrm{~V}\), obtain \(i_{s}\). Equation Transcription: Text Transcription: v_o 90 cos 2t V i_s

If \(\mathbf{V}_{o}=8 / 30^{\circ} \mathrm{V}\) in the circuit of Fig. 9.59, find \(\mathbf{I}_{s}\). Equation Transcription: Text Transcription: V_o=8 angle 30 degree V I_s

Find \(\mathbf{I}_{o}\) in the circuit of Fig. 9.60. Equation Transcription: Text Transcription: I_o

In the circuit of Fig. 9.61, find \(\mathbf{V}_{s}\) if \(\mathbf{I}_{o}=30 / 0^{\circ} \mathrm{A}\). Equation Transcription: Text Transcription: v_s I_o=30 angle 0 degree A

Find \(\mathbf{Z}\) in the network of Fig. 9.62, given that \(V_{o}=4 / 0^{\circ} \mathrm{V}\). Equation Transcription: ???? Text Transcription: ???? V_o=4 angle 0 degree V

At \(\omega=377 \mathrm{rad} / \mathrm{s}\), find the input impedance of the circuit shown in Fig. 9.63. Equation Transcription: Text Transcription: watt= 377 rad/s

At \(\omega=1 \mathrm{rad} / \mathrm{s}\), obtain the input admittance in the circuit of Fig. 9.64. Equation Transcription: Text Transcription: watt= 1 rad/s

Using Fig. 9.65, design a problem to help other students better understand impedance combinations.

For the network in Fig. 9.66, find \(Z_{\text {in }}\). Let \(\omega= 100 \mathrm{rad} / \mathrm{s}\). Equation Transcription: Text Transcription: z_in watt= 100 rad/s

Obtain \(\mathbf{Z}_{\mathrm{in}}\) for the circuit in Fig. 9.67. Equation Transcription: Text Transcription: Z_in

Find \(\mathbf{Z}_{\mathrm{eq}}\) in the circuit of Fig. 9.68. Equation Transcription: Text Transcription: Z_eq

For the circuit in Fig. 9.69, find the input impedance \(\mathbf{Z}_{\mathrm{in}}\) at \(10\ krad/s\). Equation Transcription: Text Transcription: Z_in 10 krad/s Image text transcription: For the circuit in Fig. 9.69, find the input impedance ????in at 10 krad/s.

For the circuit in Fig. 9.70, find the value of \(\mathbf{Z}_{\mathrm{T}}\). Equation Transcription: Text Transcription: Z_T

Find \(\mathbf{Z}_{\mathrm{T}}\) and \(V_{0}\) in the circuit in Fig. 9.71. Let the value of the inductance equal \(j 20 \Omega\). Equation Transcription: Text Transcription: Z_T V_o j20 ohms Image text transcription: Find ????T and Vo in the circuit in Fig. 9.71. Let the value of the inductance equal j20 ?.

Determine \(\mathbf{Z}_{\mathrm{T}}\) and \(\mathbf{I}\) for the circuit in Fig. 9.72. Equation Transcription: Text Transcription: Z_T I

For the circuit in Fig. 9.73, calculate \(\mathbf{Z}_{\mathrm{T}}\) and \(\mathbf{V}_{\mathrm{ab}}\). Equation Transcription: Text Transcription: Z_T V_ab

At \(\omega=10^{3} \mathrm{rad} / \mathrm{s}\), find the input admittance of each of the circuits in Fig. 9.74. Equation Transcription: Text Transcription: watt= 10^3 rad/s

Determine \(\mathbf{Y}_{\text {eq }}\) for the circuit in Fig. 9.75. Equation Transcription: ????eq Text Transcription: Y_eq

Find the equivalent admittance \(\mathbf{Y}_{\text {eq }}\) of the circuit in Fig. 9.76. Equation Transcription: ????eq Text Transcription: Y_eq

Find the equivalent impedance of the circuit in Fig.9.77.

Calculate the value of \(\mathbf{Z}_{\text {ab }}\) in the network of Fig. 9.79. Equation Transcription: Text Transcription: ????_ab

Calculate the value of Zab in the network of Fig.9.79.

Determine the equivalent impedance of the circuit in Fig. 9.80.

Design an \(R L\) circuit to provide a \(90^{\circ}\) leading phase shift. Equation Transcription: 90° Text Transcription: RL 90 degree

Design a circuit that will transform a sinusoidal voltage input to a cosinusoidal voltage output.

For the following pairs of signals, determine if \(v_{1}\) leads or lags \(v_{2}\) and by how much. (a) \(v_{1}=10 \cos \left(5 t-20^{\circ}\right.\) ), \(\quad v_{2}=8 \sin 5 t\) (b) \(v_{1}=19 \cos \left(2 t+90^{\circ}\right), \quad v_{2}=6 \sin 2 t\) (c) \(v_{1}=-4 \cos 10 t, \quad v_{2}=15 \sin 10 t\) Equation Transcription: Text Transcription: v_1 v_2 v_1=10 cos(5t-20 degree), v_2=8 sin 5t v_1=19 cos(2t+90 degree), v_2=6 sin 2t v_1=-4 cos 10t, v_2=15 sin 10t Image text transcription: For the following pairs of signals, determine if v1 leads or lags v2 and by how much. (a) v1 = 10 cos(5t ? 20°), v2 = 8 sin 5t (b) v1 = 19 cos(2t + 90°), v2 = 6 sin 2t (c) v1 = ?4 cos 10t, v2 = 15 sin 10t

Refer to the \(R C\) circuit in Fig. 9.81. (a) Calculate the phase shift at \(2 \mathrm{MHz}\). (b) Find the frequency where the phase shift is \(45^{\circ}\). Equation Transcription: 45° Text Transcription: RC 2 MHz 45 degree

A coil with impedance \(8+j 6 \Omega\) is connected in series with a capacitive reactance \(X\). The series combination is connected in parallel with a resistor \(R\). Given that the equivalent impedance of the resulting circuit is \(5 / 0^{\circ} \Omega\), find the value of \(R\) and \(X\). Equation Transcription: Text Transcription: 8 + j6 ohms X R 5 angle 0 degree ohms Image text transcription: A coil with impedance 8 + j6 ? is connected in series with a capacitive reactance X. The series combination is connected in parallel with a resistor R. Given that the equivalent impedance of the resulting circuit is 5? 0° ?, find the value of R and X.

(a) Calculate the phase shift of the circuit in Fig. 9.82. (b) State whether the phase shift is leading or lagging (output with respect to input). (c) Determine the magnitude of the output when the input is \(120\ V\). Equation Transcription: Text Transcription: 120 V

Consider the phase-shifting circuit in Fig. 9.83. Let \(\mathbf{V}_{i}=120 \mathrm{~V}\) operating at \(60 \mathrm{~Hz}\). Find: (a) \(\mathbf{V}_{o}\) when \(R\) is maximum (b) \(\mathbf{V}_{o}\) when \(R\) is minimum (c) the value of \(R\) that will produce a phase shift of \(45^{\circ}\) Equation Transcription: 45° Text Transcription: V_i = 120 V 60 Hz V_o R 45 degree

The ac bridge in Fig.9.37 is balanced when \(R_{1}=400 \Omega, R_{2}=600 \Omega, R_{3}=1.2 \mathrm{k} \Omega\), and \(C_{2}=0.3 \mu \mathrm{F}\). Find \(R_{x}\) and \(C_{x}\). Assume \(R_{2}\) and \(C_{2}\) are in series. Equation Transcription: Text Transcription: R_1 = 400 ohms, R_2 = 600 ohms, R_3 = 1.2 k ohms C_2 = 0.3 mu F R_x C_x R_2 C_2

A capacitance bridge balances when \(R_{1}=100 \Omega\), \(R_{2}=2 \mathrm{k} \Omega\), and \(C_{s}=40 \mu \mathrm{F}\). What is \(C_{x}\), the capacitance of the capacitor under test? Equation Transcription: Text Transcription: R_1 = 100 ohms, R_2 = 2 k ohms C_s = 40 mu F C_x

An inductive bridge balances when \(R_{1}=1.2 \mathrm{k} \Omega\), \(R_{2}=500 \Omega\), and \(L_{s}=250 \mathrm{mH}\). What is the value of \(L_{x}\), the inductance of the inductor under test? Equation Transcription: = 250 mH Text Transcription: R_1 = 1.2 k ohms, R_2 = 500 ohms L_s = 250 mH L_x

The ac bridge shown in Fig. 9.84 is known as a Maxwell bridge and is used for accurate measurement of inductance and resistance of a coil in terms of a standard capacitance \(C_{s}\). Show that when the bridge is balanced, \(L_{x}=R_{2} R_{3} C_{s} \quad \text { and } \quad R_{x}=\frac{R_{2}}{R_{1}} R_{3}\) Find \(L_{x}\) and \(R_{x}\) for \(R_{1}=40 \mathrm{k} \Omega, R_{2}=1.6 \mathrm{k} \Omega\), \(R_{3}=4 \mathrm{k} \Omega\), and \(C_{s}=0.45 \mu \mathrm{F}\). Equation Transcription: Text Transcription: C_s L_x=R_2R_3C_s R_x=R_2/R_1 R3 L_x R_x R_1=40 k ohms, R_2=1.6 k ohms, R_3=4 k ohms C_s=0.45 mu F

The ac bridge circuit of Fig. 9.85 is called a . It is used for measuring the frequency of a source. Show that when the bridge is balanced, \(f=\frac{1}{2 \pi \sqrt{R_{2} R_{4} C_{2} C_{4}}}\) Equation Transcription: Text Transcription: f=1/2pi square root R_2 R_4 C_2 C_4

The circuit shown in Fig. 9.86 is used in a television receiver. What is the total impedance of this circuit?

The network in Fig. 9.87 is part of the schematic describing an industrial electronic sensing device. What is the total impedance of the circuit at \(4\ kHz\)? Equation Transcription: Text Transcription: 4 kHz

A series audio circuit is shown in Fig. 9.88. (a) What is the impedance of the circuit? (b) If the frequency were halved, what would be the impedance of the circuit?

An industrial load is modeled as a series combination of an inductor and a resistance as shown in Fig. 9.89. Calculate the value of a capacitor \(C\) across the series combination so that the net impedance is resistive at a frequency of \(2\ kHz\). Equation Transcription: Text Transcription: C 2 kHz

An industrial coil is modeled as a series combination of an inductance \(L\) and resistance \(R\), as shown in Fig. 9.90. Since an ac voltmeter measures only the magnitude of a sinusoid, the following measurements are taken at \(60 \mathrm{~Hz}\) when the circuit operates in the steady state: \(\left|\mathbf{V}_{s}\right|=145 \mathrm{~V}, \quad\left|\mathbf{V}_{1}\right|=50 \mathrm{~V}, \quad\left|\mathbf{V}_{o}\right|=110 \mathrm{~V}\) Use these measurements to determine the values of \(L\) and \(R\). Equation Transcription: Text Transcription: L R 60 Hz |V_s|=145 V, |V_1|=50 V, |V_0|=110 V

Figure 9.91 shows a series combination of an inductance and a resistance. If it is desired to connect a capacitor in parallel with the series combination such that the net impedance is resistive at \(10\ kHz\), what is the required value of \(C\)? Equation Transcription: Text Transcription: 10 kHz C

A transmission line has a series impedance of \(\mathbf{Z}=100 / 75^{\circ} \Omega\) and a shunt admittance of \(\mathbf{Y}=\) \(450 / 48^{\circ} \mu \mathrm{S}\). Find: (a) the characteristic impedance \(\mathbf{Z}_{o}=\sqrt{\mathbf{Z} / \mathbf{Y}},(\mathrm{b})\) the propagation constant \(\gamma=\sqrt{\mathbf{Z Y}}\). Equation Transcription: Z= ???? = Text Transcription: Z=100 angle 75 degree ohms ???? =450 angle 48 degree mu S Square root Z/Y gamma=Square root Z/Y Image text transcription: A transmission line has a series impedance of Z = 100? 75° ? and a shunt admittance of Y = 450? 48° ?S. Find: (a) the characteristic impedance Zo = ? _____ Z?Y , (b) the propagation constant ? = ? ___ ZY .

A power transmission system is modeled as shown in Fig. 9.92. Given the source voltage and circuit elements \(\mathbf{V}_{s}=115 / 0^{\circ} \mathrm{V}, \quad\) source impedance \(Z_{s}=(1+j 0.5) \Omega, \quad\) line impedance \(\mathbf{Z}_{t}=(0.4+j 0.3) \Omega, \quad\) and load impedance \(Z_{L}=(23.2+j 18.9) \Omega, \quad\) find the load current \(\mathbf{I}_{L}\). Equation Transcription: Text Transcription: V_s=115 angle 0 degree V Z_s=(1+j0.5) ohms Z_t=(0.4+j0.3) ohms Z_L=(23.2+j18.9) ohms I_L