Delmar's Standard Textbook Of Electricity
7th Edition
ISBN: 9781337900348
Author: Stephen L. Herman
Publisher: Cengage Learning
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Chapter 30 Solutions
Delmar's Standard Textbook Of Electricity
Ch. 30 - Prob. 1RQCh. 30 - Prob. 2RQCh. 30 - Prob. 3RQCh. 30 - Prob. 4RQCh. 30 - What type of motor is known as a constant-speed...Ch. 30 - What is CEMF?Ch. 30 - Prob. 7RQCh. 30 - Prob. 8RQCh. 30 - Prob. 9RQCh. 30 - What type of motor should never be operated at no...
Ch. 30 - Prob. 11RQCh. 30 - What is the most common way of changing the...Ch. 30 - Prob. 13RQCh. 30 - Prob. 14RQCh. 30 - What device is used to disconnect power to the...Ch. 30 - Prob. 16RQCh. 30 - Who was the first person to establish a...Ch. 30 - One horsepower is equal to how many W?Ch. 30 - A motor is operating a load that requires a torque...Ch. 30 - The motor in Question 19 is connected to a 250-VDC...Ch. 30 - Prob. 1PA
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- A17)arrow_forwardUsing Carson's rule, determine the transmission bandwidth for commercial FM radio broadcasting, provided that the maximum value of frequency deviation is 75 kHz and the bandwidth of the audio signal is 15 kHzarrow_forward2. Laboratory Preliminary Discussion First-order High-pass RC Filter Analysis The first-order high-pass RC filter shown in figure 3 below represents all voltages and currents in the time domain. We will again convert the circuit to its s-domain equivalent as shown in figure 4 and apply Laplace transform techniques. ic(t) C vs(t) i₁(t) + + vc(t) R1 ww Vi(t) || 12(t) V2(t) R₂ Vout(t) VR2(t) = V2(t) Figure 3: A first-order high-pass RC filter represented in the time domain. Ic(s) C + Vs(s) I₁(s) + + Vc(s) R₁ www V₁(s) 12(s) V₂(s) R₂ Vout(S) = VR2(S) = V2(s) Figure 4: A first-order high-pass RC filter represented in the s-domain. Again, to generate the s-domain expression for the output voltage, You (S) = V2 (s), for the circuit shown in figure 4 above, we can apply voltage division in the s-domain as shown in equation 2 below. Equation 2 will be used in the prelab computations to find an expression for the output voltage, xc(t), in the time domain. equation (2) R₂ Vout(s) = V₂(s) = R₂+…arrow_forward
- Can you show me the steps to get the last part after the second equal sign.arrow_forwardPrelab Information 1. Laboratory Preliminary Discussion First-order Low-pass RC Filter Analysis The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques. vs(t) i₁(t) + R₁ ww V₁(t) 12(t) Lic(t) Vout(t) = V2(t) R₂ Vc(t) C Vc(t) VR2(t) = V2(t) + Vs(s) Figure 1: A first-order low-pass RC filter represented in the time domain. I₁(s) R1 W + V₁(s) V₂(s) 12(s) Ic(s) + Vout(S) == Vc(s) Vc(s) Zc(s) = = VR2(S) V2(s) Figure 2: A first-order low-pass RC filter represented in the s-domain.arrow_forwardA.15 Consider a communication channel, transfer characteristic of which is defined by the nonlinear relation, y(t) = x(t) + x² (t), where x(t) is the input and y(t) is the output. Assuming the input is an FM signal, x(t) = cos (2лft+(t)), find y(t). Is it possible to retrieve x(t) from y(t)? If so, how?arrow_forward
- 1) Show that a regenerative receiver can be used to recover message from the following modulated signals. a. DSB-PC b. DSB-SC 1b) Does the receiver need to recover the carrier phase? 1c) What are the filtering requirements and restrictions on message signal bandwidth and carrier frequency.arrow_forward2) Estimate the transmission bandwidth for the following FM modulated signals (W is the message bandwidth) a) W1KHz and frequency deviation of 75KHz b) W = 20KHz and frequency deviation of 75KHz c) W1KHz and frequency deviation of 150KHz d) W20KHz and frequency deviation of 150KHZarrow_forwardI want to explain how the result becomes (735.1) Hz) and what are the steps and explain the reasons? Q6 The FET shown in Fig. 1.43 has gm = 3.4mS and ra =100 K. Find the approximate lower cutoff frequency. Ans: 735.1 Hz. 25V 2ΚΩ 1.5ΜΩ 0.02µF 0.02µF 20 ΚΩ 330kQ 820 ΩΣ OpF Fig. 1.43 Circuit for Q6. 40ΚΩarrow_forward
- 3. What is the function of LM565 pin 6? 4. What is the purpose of the multistage low-pass filter between the LM565 output and the comparator input? C10.1μ FSK Input w₁ R2 100k -o+5V(Vcc) VR1 10k C4 C5: 0.1 μ. 0.1μ 0.1 μ 8 10 R3 R4 D₁ FSK Phase Rx 7 10K 10K Detector www ww ww 1N4004 + Demodulated Output 6 AMP R₁ 6 100k 3 C₂ 0.05 μ VCO 4 5 9 U1 -5V LM565 -0-5V(VEE) Fig. 14-2 FSK demodulator U2 R6 μ4741 10karrow_forward1. What components determine the free-running frequency of the VCO in LM565 of Fig. 14-2? 2. What is the purpose of μA741 in Fig. 14-2? C10.1μ FSK Input -o+5V(Vcc) VR1 10k C4 C5: 0.1 μ. 0.1 μ 0.1 μ 8 10 R3 R4 R5 Phase Rx 7 10K 10K 10k D₁ FSK Detector www ww ww ww 1N4004 + Demodulated Output AMP 6 R₁ 6 100k w₁ R2 100k 3 C₂ 0.05 μ VCO 4 5 9 U1 -5V LM565 -0-5V(VEE) Fig. 14-2 FSK demodulator U2 R6 μ4741 10karrow_forwardWhen troubleshooting power and control circuits, approximate meter readings should be anticipated if the meter readings are going to be used to help determine circuit problems. Determine the expected DMM reading if the ciircuit is working properly. The expected reading of DMM 1 with the motor on is what VAC? And the expected reading of DMM 2 with the motor is on is what VAC? And The expected reading of DMM 3 with the motor on is What mA?arrow_forward
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