7. You want to build a circuit which causes a ligh tbulb to turn on when you throw a switch. So, you build the circuit if Figure 5. When the switch is connected to point A the capacitor charges, and when you connect the switch to point B the light goes on. Let the resistance of the light bulb be Rℓ = 1.50 ×103 Ω, the potential difference across the battery is VB = 10.0V , and the capacitor has a capacitance of C = 1.35 × 10−4 F
75uS is the time constant used for a high-pass filter to enhance the high frequency audio before transmission to help reduce noise upon reception. The PLL is the portion of the IC which locks your chosen transmission frequency to the crystal reference X1. The PLL portion also contains an oscillator circuit which works in conjunction with the external parts of D5 and the STUB (that weird trace on the back of the board). D5 is called a varactor diode, and is a special variety of diode that is connected backwards. As a reverse DC voltage is applied across the diode, its capacitance varies.
Apply this bias to all micromanometer measurements. 2. Connect the micromanometer, pressure transducer (voltmeter), and scannivalve to the Pitot tube in the subsonic tunnel. 3. Record the readings of the three instruments at eight different speed settings of the tunnel: 15, 20, 25, 30, 35, 40, 45, & 50.
Trigonometry Problem #1 The electricity supplied to your house is called alternating current (AC) because the current varies sinusoidally with time. The voltage which causes the current to flow also varies sinusoidally with time. Bothe the current and the voltage have a frequency of 60 cycles per second which corresponds to a period of 1/60 of a second, but current and voltage have different phase shifts. We will be determining equations for the voltages and the current as functions of time based upon the equationf(t)=Acos[B(x-C)]+D. Under the assumption that current and voltage are both sinusoidal functions of time (t, in seconds) and both have the same frequency of 60 cycles per second (period=1/60), let C be the current in Amperes (Amps) with a maximum current of 5 Amps at zero seconds (t=zero).
The spectrophotometer was set for 360 nm and the mode was changed to transmittance. The zero adjustment knob and 100% adjustment knob (using the blank) adjusted accordingly. The absorbance of each sample was measured at 360 nm. The wavelength was then set to 380 nm. The absorbance of each sample was calculated again after fixing the 100% adjustment knob to read 100.
Calculate the impedance by the measured values of voltage and current from the oscilloscope. Also calculate the impedance from = − C. Answer: Impedance, Z = 5∠0 34.65 60.48°) = 173.25 - 60.48⁰ = 85.37 Ω - j150.76 Ω Again, XC = 1/ = 1/ (6280 x 10-6) = 159.24
A 17. D 18. A and D 19. C 20. B, C and D Electrical Circuit – an electrical device that provides a path for electrical current to flow Frequency – The cycles per second of alternating current, measured in Hertz Amplitude – the maximum difference of an alternating electric current or potential from the average value Phrase – is a position of a point in time on a waveform cycle Bit time – the time it takes for one bit to be ejected from a network interface card at a standard speed Encoding Scheme – a system used by computers to represent data as characters Unshielded twisted-pair – The word unshielded refers to the fact that UTP cables have no added shielding materials to prevent EMI problems Core- The light source on one end of the cable shines light into the core Cladding- surrounds the core, for the entire length of the cable, and reflects the light into the core Radio waves- work well for networking because as a waveform, radio waves can be changed (modulated) over time to send data.
They are Amplitude modulation (AM), Frequency Modulation (FM), and Phase Modulation (PM). These methods are based on altering one of the carrier’s characteristics: amplitude, frequency, or phase. In AM, FM and PM modulations each carrier changes in accordance with the information signal’s amplitude variations. When it comes to advantages and disadvantages there are performance factors to consider. They are spectral efficiency which identifies a communication system’s ability to achieve a given data rate within a given bandwidth.
CIS 312 Final Exam Answers https://hwguiders.com/downloads/cis-312-final-exam-answers CIS 312 Final Exam Answers • Question 1 A(n) ____ increases a signal’s strength (amplitude) and can extend a signal’s range by boosting signal power to overcome attenuation. • Question 2 Square waves can be generated by rapidly switching (pulsing) an electrical or optical power source—a technique called ____. • Question 3 A(n) ____ is the maximum number of bits or bytes per second that the channel can carry. • Question 4 Messages can be loosely categorized into two types—data messages and command messages. • Question 5 ____ mode uses a single shared channel, and each node takes turns using the transmission line to transmit and receive.
c. The circuit creates a direct current. d. The circuit creates an alternating current. 2. A PC NIC and a switch port create one electrical circuit to use when sending data. The sender creates a (maximum) 1-volt electrical signal with a frequency of 1000 hertz.