Principles Of Electric Circuits
10th Edition
ISBN: 9780134879482
Author: Floyd, Thomas L.
Publisher: Pearson,
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Chapter 3, Problem 9TFQ
Ohm’s law for finding voltage is V = I/R.
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Chapter 3 Solutions
Principles Of Electric Circuits
Ch. 3 - If the current drops to 10 mA under the same...Ch. 3 - If the sense resistor develops 0.8 V across it,...Ch. 3 - Calculate the current in Figure 3-71 if R is...Ch. 3 - What is the current in mA produced by 1 kV across...Ch. 3 - How much current is there through a 6.8 M resistor...Ch. 3 - In Figure 3-10, how much voltage is required to...Ch. 3 - lf there are 3.2 A through a 47 resistor, what is...Ch. 3 - If there are 450 A through a 3.9 M resistor, what...Ch. 3 - In the circuit of Figure 3-13, how much resistance...Ch. 3 - If one of the grid wires opens, the current drops...
Ch. 3 - If the resistor is changed in Figure 3-14 so that...Ch. 3 - If the total resistance of a circuit increases and...Ch. 3 - Ohms law for finding resistance is R = I/V.Ch. 3 - When milliamps and kilohms are multiplied...Ch. 3 - If a 10 k resistor is connected to a 10 V source,...Ch. 3 - The current in a fixed resistor is directly...Ch. 3 - Ohms law for finding current is I = V/R.Ch. 3 - When microamps and megohms are multiplied, the...Ch. 3 - When voltage is constant, current is inversely...Ch. 3 - Ohms law for finding voltage is V = I/R.Ch. 3 - When I is plotted as a function of V for a fixed...Ch. 3 - Ohms law states that 1. current equals voltage...Ch. 3 - When the voltage across a resistor is doubled, the...Ch. 3 - When 10 V are applied across a 20 resistor, the...Ch. 3 - When there are 10 mA of current through 1.0 k...Ch. 3 - If 20 V are applied across a resistor and there...Ch. 3 - A current of 250 A through a 4.7 k resistor...Ch. 3 - A resistance of 2.2 M is connected across a 1 kV...Ch. 3 - How much resistance is required to limit the...Ch. 3 - An electric heater draws 2.5 A from a 110 V...Ch. 3 - The current through a flashlight bulb is 20 mA and...Ch. 3 - If the current through a fixed resistor goes from...Ch. 3 - If the voltage across a fixed resistor goes from...Ch. 3 - A variable resistor has 5 V across it. If you...Ch. 3 - If the voltage across a resistor increases from 5...Ch. 3 - If larger voltages are applied and results are...Ch. 3 - If the IV curve for a larger value resistor is...Ch. 3 - If the voltmeter reading changes to 175 V, the...Ch. 3 - If is changed to a larger value and the voltmeter...Ch. 3 - If the resistor is removed from the circuit...Ch. 3 - If the resistor is removed from the circuit...Ch. 3 - If the rheostat is adjusted to increase the...Ch. 3 - If the rheostat is adjusted to increase the...Ch. 3 - If the fuse opens, the voltage across the heating...Ch. 3 - If the source voltage increases, the voltage...Ch. 3 - If the fuse is changed to one with a higher...Ch. 3 - If the lamp burns out (opens), the current a....Ch. 3 - If the lamp burns out, the voltage across it a....Ch. 3 - In a circuit consisting of a voltage source and a...Ch. 3 - State the formula used to find I when the values...Ch. 3 - State the formula used to find V when the values...Ch. 3 - State the formula used to find R when the values...Ch. 3 - A variable voltage source is connected to the...Ch. 3 - In a certain circuit, I = 5 mA when V = 1 V....Ch. 3 - Figure 322 is a graph of current versus voltage...Ch. 3 - Plot the currentvoltage relationship for a...Ch. 3 - Plot the currentvoltage relationship for a...Ch. 3 - Determine the current in each circuit in Figure...Ch. 3 - You are measuring the current in a circuit that is...Ch. 3 - (a) If you wish to increase the amount of current...Ch. 3 - Plot a graph of current versus voltage for voltage...Ch. 3 - Does the graph in Problem 13 indicate a linear...Ch. 3 - Figure 3-24 shows an IV curve for a certain light...Ch. 3 - For the bulb graphed in Figure 3-24, what is the...Ch. 3 - Determine the current in each case: a. V = 5 V, R...Ch. 3 - Determine the current in each case: a. V = 9 V, R...Ch. 3 - Assume 200 mV is across a 330 m current sensing...Ch. 3 - A certain resistor has the following color code:...Ch. 3 - A 4-band resistor is connected across the...Ch. 3 - A 5-band resistor is connected across a 12 V...Ch. 3 - If the voltage in Problem 22 is doubled, will a...Ch. 3 - A certain rear window defroster has a resistance...Ch. 3 - If the voltage of the battery in problem 24 drops...Ch. 3 - The potentiometer connected as a rheostat in...Ch. 3 - A 270 current-limiting resistor has a voltage of...Ch. 3 - A small solar cell is connected to a 27 k...Ch. 3 - Calculate the voltage for each value of I and R:...Ch. 3 - Calculate the voltage for each value of I and R:...Ch. 3 - Three amperes of current are measured through a 27...Ch. 3 - Assign a voltage value to each source in the...Ch. 3 - A 6 V source is connected to a 100 resistor by...Ch. 3 - Calculate the resistance of a rheostat for each...Ch. 3 - Calculate the resistance of a rheostat for each...Ch. 3 - Six volts is applied across a resistor. A current...Ch. 3 - The filament of a lamp in the circuit of Figure...Ch. 3 - A certain electrical device has an unknown...Ch. 3 - By varying the rheostat (variable resistor) in the...Ch. 3 - In the light circuit of Figure 329, identify the...Ch. 3 - Assume you have a 32-light string and one of the...
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- Imaginary Axis (seconds) 1 6. Root locus for a closed-loop system with L(s) = is shown below. s(s+4)(s+6) 15 10- 0.89 0.95 0.988 0.988 -10 0.95 -15 -25 0.89 20 Root Locus 0.81 0.7 0.56 0.38 0.2 5 10 15 System: sys Gain: 239 Pole: -0.00417 +4.89 Damping: 0.000854 Overshoot (%): 99.7 Frequency (rad/s): 4.89 System: sys Gain: 16.9 Pole: -1.57 Damping: 1 Overshoot (%): 0 Frequency (rad/s): 1.57 0.81 0.7 0.56 0.38 0.2 -20 -15 -10 -5 5 10 Real Axis (seconds) From the values shown in the figure, compute the following. a) Range of K for which the closed-loop system is stable. b) Range of K for which the closed-loop step response will not have any overshoot. Note that when all poles are real, the step response has no overshoot. c) Smallest possible peak time of the system. Note that peak time is the smallest when wa is the largest for the dominant pole. d) Smallest possible settling time of the system. Note that peak time is the smallest when σ is the largest for the dominant pole.arrow_forwardFor a band-rejection filter, the response drops below this half power point at two locations as visualised in Figure 7, we need to find these frequencies. Let's call the lower frequency-3dB point as fr and the higher frequency -3dB point fH. We can then find out the bandwidth as f=fHfL, as illustrated in Figure 7. 0dB Af -3 dB Figure 7. Band reject filter response diagram Considering your AC simulation frequency response and referring to Figure 7, measure the following from your AC simulation. 1% accuracy: (a) Upper-3db Frequency (fH) = Hz (b) Lower-3db Frequency (fL) = Hz (c) Bandwidth (Aƒ) = Hz (d) Quality Factor (Q) =arrow_forwardP 4.4-21 Determine the values of the node voltages V1, V2, and v3 for the circuit shown in Figure P 4.4-21. 29 ww 12 V +51 Aia ww 22. +21 ΖΩ www ΖΩ w +371 ①1 1 Aarrow_forward
- 1. What is the theoretical attenuation of the output voltage at the resonant frequency? Answer to within 1%, or enter 0, or infinity (as “inf”) Attenuation =arrow_forwardWhat is the settling time for your output signal (BRF_OUT)? For this question, We define the settling time as the period of time it has taken for the output to settle into a steady state - ie when your oscillation first decays (aka reduces) to less than approximately 1/20 (5%) of the initial value. (a) Settling time = 22 μs Your last answer was interpreted as follows: Incorrect answer. Check 22 222 What is the peak to peak output voltage (BRF_OUT pp) at the steady state condition? You may need to use the zoom function to perform this calculation. Select a time point that is two times the settling time you answered in the question above. Answer to within 10% accuracy. (a) BRF_OUT pp= mVpp As you may have noticed, the output voltage amplitude is a tiny fraction of the input voltage, i.e. it has been significantly attenuated. Calculate the attenuation (decibels = dB) in the output signal as compared to the input based on the formula given below. Answer to within 1% accuracy.…arrow_forwardmy previous answers for a,b,d were wrong a = 1050 b = 950 d=9.99 c was the only correct value i got previously c = 100hz is correctarrow_forward
- V₁(t) ww ZRI ZLI ZL2 ZTH Zci VTH Zc21 Figure 8. Circuit diagram showing calculation approach for VTH and Z TH we want to create a blackbox for the red region, we want to use the same input signal conditions as previously the design of your interference ector circuit: Sine wave with a 1 Vpp, with a frequency of 100 kHz (interference) Square wave with 2.4Vpp, with a frequency of 10 kHz (signal) member an AC Thevenin equivalent is only valid at one frequency. We have chosen to calculate the Thevenin equivalent circuit (and therefore the ackbox) at the interference frequency (i.e. 100 kHz), and the signal frequency (i.e. 10 kHz) as these are the key frequencies to analyse. Your boss is assured you that the waveform converter module has been pre-optimised to the DAB Receiver if you use the recommended circuit topology.arrow_forwardVs(t) + v(t) + vi(t) ZR ZL Figure 1: Second order RLC circuit Zc + ve(t) You are requested to design the circuit shown in Figure 1. The circuit is assumed to be operating at its resonant frequency when it is fed by a sinusoidal voltage source Vs (t) = 2sin(le6t). To help design your circuit you have been given the value of inductive reactance ZL = j1000. Assume that the amplitude of the current at resonance is Is (t) = 2 mA. Based on this information, answer the following to help design your circuit. Use cartesian notation for your answers, where required.arrow_forwardWhat is the attenuation at the resonant frequency? You should use the LTSpice cursors for your measurement. Answer to within 1% accuracy, or enter 0, or infinity (as "inf") (a) Attenuation (dB) = dB Check You may have noticed that it was significantly easier to use frequency-domain "AC" simulation to measure the attenuation, compared to the steps we performed in the last few questions. (i.e. via a time-domain "transient" simulation). AC analysis allows us to observe and quantify large scale positive or negative changes in a signal of interest across a wide range of different frequencies. From the response you will notice that only frequencies that are relatively close to 100 kHz have been attenuated. This is the result of the Band-reject filter you have designed, and shows the 'rejection' (aka attenuation) of any frequencies that lie in a given band. The obvious follow-up question is how do we define this band? We use a quantity known as the bandwidth. A commonly used measurement for…arrow_forward
- V₁(t) ww ZRI ZLI ZL2 ZTH Zci VTH Zc21 Figure 8. Circuit diagram showing calculation approach for VTH and Z TH we want to create a blackbox for the red region, we want to use the same input signal conditions as previously the design of your interference ector circuit: Sine wave with a 1 Vpp, with a frequency of 100 kHz (interference) Square wave with 2.4Vpp, with a frequency of 10 kHz (signal) member an AC Thevenin equivalent is only valid at one frequency. We have chosen to calculate the Thevenin equivalent circuit (and therefore the ackbox) at the interference frequency (i.e. 100 kHz), and the signal frequency (i.e. 10 kHz) as these are the key frequencies to analyse. Your boss is assured you that the waveform converter module has been pre-optimised to the DAB Receiver if you use the recommended circuit topology.arrow_forwardVs(t) + v(t) + vi(t) ZR ZL Figure 1: Second order RLC circuit Zc + ve(t) You are requested to design the circuit shown in Figure 1. The circuit is assumed to be operating at its resonant frequency when it is fed by a sinusoidal voltage source Vs (t) = 2sin(le6t). To help design your circuit you have been given the value of inductive reactance ZL = j1000. Assume that the amplitude of the current at resonance is Is (t) = 2 mA. Based on this information, answer the following to help design your circuit. Use cartesian notation for your answers, where required.arrow_forwardFor a band-rejection filter, the response drops below this half power point at two locations as visualised in Figure 7, we need to find these frequencies. Let's call the lower frequency-3dB point as fr and the higher frequency -3dB point fH. We can then find out the bandwidth as f=fHfL, as illustrated in Figure 7. 0dB Af -3 dB Figure 7. Band reject filter response diagram Considering your AC simulation frequency response and referring to Figure 7, measure the following from your AC simulation. 1% accuracy: (a) Upper-3db Frequency (fH) = Hz (b) Lower-3db Frequency (fL) = Hz (c) Bandwidth (Aƒ) = Hz (d) Quality Factor (Q) =arrow_forward
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