General/Chapter 4 Study Guide - Revision history

Admin: /* Section 4.2: AC Power */

2016-05-28T01:36:20Z

Section 4.2: AC Power

← Older revision		Revision as of 01:36, 28 May 2016
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	* The RMS voltage across a 50 Ohm dummy load dissipating 1200 W is <math> P = \frac{V_{RMS}^2}{R}</math> so <math>E_{RMS} = \sqrt{PR} = 245 V</math>		* The RMS voltage across a 50 Ohm dummy load dissipating 1200 W is <math> P = \frac{V_{RMS}^2}{R}</math> so <math>E_{RMS} = \sqrt{PR} = 245 V</math>
	* If average power is measured as 1060 watts for an unmodulated carrier, its PEP output is 1060 watts. For unmodulated signals, PEP = average power		* If average power is measured as 1060 watts for an unmodulated carrier, its PEP output is 1060 watts. For unmodulated signals, PEP = average power
	* If oscilloscope measures 500 V peak-to-peak across 50 ohm load, PEP is <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{(0.707 \frac{V_{PK-TO-PK}}{2})^2}{R}</math>		* If oscilloscope measures 500 V peak-to-peak across 50 ohm load, PEP is <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{(0.707 \frac{V_{PK-TO-PK}}{2})^2}{R} = 625 W</math>

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Section 4.2: AC Power */

2016-05-28T01:35:15Z

Section 4.2: AC Power

← Older revision		Revision as of 01:35, 28 May 2016
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	* The RMS voltage across a 50 Ohm dummy load dissipating 1200 W is <math> P = \frac{V_{RMS}^2}{R}</math> so <math>E_{RMS} = \sqrt{PR} = 245 V</math>		* The RMS voltage across a 50 Ohm dummy load dissipating 1200 W is <math> P = \frac{V_{RMS}^2}{R}</math> so <math>E_{RMS} = \sqrt{PR} = 245 V</math>
	* If average power is measured as 1060 watts for an unmodulated carrier, its PEP output is 1060 watts. For unmodulated signals, PEP = average power		* If average power is measured as 1060 watts for an unmodulated carrier, its PEP output is 1060 watts. For unmodulated signals, PEP = average power
	* If oscilloscope measures 500 V peak-to-peak across 50 ohm load, PEP is <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{0.707 \frac{V_{PK-TO-PK}}{2})^2}{R}</math>		* If oscilloscope measures 500 V peak-to-peak across 50 ohm load, PEP is <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{(0.707 \frac{V_{PK-TO-PK}}{2})^2}{R}</math>

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Section 4.2: AC Power */

2016-05-28T01:35:04Z

Section 4.2: AC Power

← Older revision		Revision as of 01:35, 28 May 2016
Line 18:		Line 18:
	* The RMS voltage across a 50 Ohm dummy load dissipating 1200 W is <math> P = \frac{V_{RMS}^2}{R}</math> so <math>E_{RMS} = \sqrt{PR} = 245 V</math>		* The RMS voltage across a 50 Ohm dummy load dissipating 1200 W is <math> P = \frac{V_{RMS}^2}{R}</math> so <math>E_{RMS} = \sqrt{PR} = 245 V</math>
	* If average power is measured as 1060 watts for an unmodulated carrier, its PEP output is 1060 watts. For unmodulated signals, PEP = average power		* If average power is measured as 1060 watts for an unmodulated carrier, its PEP output is 1060 watts. For unmodulated signals, PEP = average power
	* If oscilloscope measures 500 V peak-to-peak across 50 ohm load, PEP is <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{0.707 \frac{V_{PK-TO-PK}}{2}})^2}{R}</math>		* If oscilloscope measures 500 V peak-to-peak across 50 ohm load, PEP is <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{0.707 \frac{V_{PK-TO-PK}}{2})^2}{R}</math>

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Section 4.2: AC Power */

2016-05-28T01:34:49Z

Section 4.2: AC Power

← Older revision		Revision as of 01:34, 28 May 2016
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	* oscilloscope measures 200 V peak-to-peak across 50 Ohm dummy load. What is PEP output? <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{(0.707 V_{PK})^2}{R} = \dfrac{(0.707 \frac{V_{PK-TO-PK}}{2})^2}{R} = 245 V</math>		* oscilloscope measures 200 V peak-to-peak across 50 Ohm dummy load. What is PEP output? <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{(0.707 V_{PK})^2}{R} = \dfrac{(0.707 \frac{V_{PK-TO-PK}}{2})^2}{R} = 245 V</math>
			* AC signal producing same power dissipation in a resistor as a DC signal of the same voltage is the AC signal with an RMS voltage equal to the DC voltage: <math>V_{RMS,AC} = V_{DC}</math>
			* A sine wave with a peak voltage <math>V_{PK} = 17 \text{V}</math> has an RMS voltage of <math>V_{RMS} = 0.707 V_{PK} = 12 \text{V}</math>
			* For an unmodulated carrier, PEP = average power
			* The RMS voltage across a 50 Ohm dummy load dissipating 1200 W is <math> P = \frac{V_{RMS}^2}{R}</math> so <math>E_{RMS} = \sqrt{PR} = 245 V</math>
			* If average power is measured as 1060 watts for an unmodulated carrier, its PEP output is 1060 watts. For unmodulated signals, PEP = average power
			* If oscilloscope measures 500 V peak-to-peak across 50 ohm load, PEP is <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{0.707 \frac{V_{PK-TO-PK}}{2}})^2}{R}</math>

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Section 4.1: Current, Voltage, Power */

2016-05-28T01:31:22Z

Section 4.1: Current, Voltage, Power

← Older revision		Revision as of 01:31, 28 May 2016
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	==Section 4.1: Current, Voltage, Power==		==Section 4.1: Current, Voltage, Power==

	* an ~~incresae~~ in power of 2x is equal to 3 dB		* an increase in power of 2x is equal to 3 dB
	* in a purely resistive parallel circuit, the total amount of current is the sum of each branch current		* in a purely resistive parallel circuit, the total amount of current is the sum of each branch current
	* if 400 V dc is supplied to an 800 ohm load, use the formula <math>P = \dfrac{E^2}{R} = \dfrac{400^2}{800} = 200 \text{W}</math>		* if 400 V dc is supplied to an 800 ohm load, use the formula <math>P = \dfrac{E^2}{R} = \dfrac{400^2}{800} = 200 \text{W}</math>
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	* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 \text{mW}</math>		* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 \text{mW}</math>
	* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 \log(PR)</math>, which rearranges to <math>PR = 10^{-\frac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%		* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 \log(PR)</math>, which rearranges to <math>PR = 10^{-\frac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%

			==Section 4.2: AC Power==

			* oscilloscope measures 200 V peak-to-peak across 50 Ohm dummy load. What is PEP output? <math>PEP = \dfrac{E_{RMS}^2}{R} = \dfrac{(0.707 V_{PK})^2}{R} = \dfrac{(0.707 \frac{V_{PK-TO-PK}}{2})^2}{R} = 245 V</math>

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Section 4.1: Current, Voltage, Power */

2016-05-28T01:27:41Z

Section 4.1: Current, Voltage, Power

← Older revision		Revision as of 01:27, 28 May 2016
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	* if a 12 V DC light bulb draws 0.2 A, use the formula <math>P = I E = (12)(0.2) = 2.4 \text{W}</math>		* if a 12 V DC light bulb draws 0.2 A, use the formula <math>P = I E = (12)(0.2) = 2.4 \text{W}</math>
	* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 \text{mW}</math>		* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 \text{mW}</math>
	* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 log(PR)</math>, which rearranges to <math>PR = 10^{-\frac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%		* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 \log(PR)</math>, which rearranges to <math>PR = 10^{-\frac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Section 4.1: Current, Voltage, Power */

2016-05-28T01:27:28Z

Section 4.1: Current, Voltage, Power

← Older revision		Revision as of 01:27, 28 May 2016
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	* an incresae in power of 2x is equal to 3 dB		* an incresae in power of 2x is equal to 3 dB
	* in a purely resistive parallel circuit, the total amount of current is the sum of each branch current		* in a purely resistive parallel circuit, the total amount of current is the sum of each branch current
	* if 400 V dc is supplied to an 800 ohm load, use the formula <math>P = \dfrac{E^2}{R} = \dfrac{400^2}{800} = 200 W</math>		* if 400 V dc is supplied to an 800 ohm load, use the formula <math>P = \dfrac{E^2}{R} = \dfrac{400^2}{800} = 200 \text{W}</math>
	* if a 12 V DC light bulb draws 0.2 A, use the formula <math>P = I E = (12)(0.2) = 2.4 W</math>		* if a 12 V DC light bulb draws 0.2 A, use the formula <math>P = I E = (12)(0.2) = 2.4 \text{W}</math>
	* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 mW</math>		* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 \text{mW}</math>
	* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 log(PR)</math>, which rearranges to <math>PR = 10^{-\frac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%		* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 log(PR)</math>, which rearranges to <math>PR = 10^{-\frac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%

Admin: /* Section 4.1: Current, Voltage, Power */

2016-05-28T01:27:00Z

Section 4.1: Current, Voltage, Power

← Older revision		Revision as of 01:27, 28 May 2016
Line 8:		Line 8:
	* if a 12 V DC light bulb draws 0.2 A, use the formula <math>P = I E = (12)(0.2) = 2.4 W</math>		* if a 12 V DC light bulb draws 0.2 A, use the formula <math>P = I E = (12)(0.2) = 2.4 W</math>
	* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 mW</math>		* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 mW</math>
	* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 log(PR)</math>, which rearranges to <math>PR = 10^{-\~~dfrac~~{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%		* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 log(PR)</math>, which rearranges to <math>PR = 10^{-\frac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Chapter 4: Components and Circuits */

2016-05-28T01:26:42Z

Chapter 4: Components and Circuits

← Older revision		Revision as of 01:26, 28 May 2016
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	=Chapter 4: Components and Circuits=		=Chapter 4: Components and Circuits=

	(~~Fill in~~)		==Section 4.1: Current, Voltage, Power==

			* an incresae in power of 2x is equal to 3 dB
			* in a purely resistive parallel circuit, the total amount of current is the sum of each branch current
			* if 400 V dc is supplied to an 800 ohm load, use the formula <math>P = \dfrac{E^2}{R} = \dfrac{400^2}{800} = 200 W</math>
			* if a 12 V DC light bulb draws 0.2 A, use the formula <math>P = I E = (12)(0.2) = 2.4 W</math>
			* if 7 mA flows through 1.25 kOhm load, the amount of power dissipated can be found using <math>P = I^2 R = (7.0 \times 10^{-3})^2 (1.25 \times 10^3) = 61 mW</math>
			* a power transmission line loss of 1 dB: to find percent power loss, use the formula <math>dB = 10 log(PR)</math>, which rearranges to <math>PR = 10^{-\dfrac{1}{10}} = .794</math> so the power is 79.4% of what it was. This corresponds to a loss of <math>100 - 79.4 = 20.6</math> or 20.6%

	==Section 4.3: Basic Components==		==Section 4.3: Basic Components==

Admin: /* Flags */

2016-05-28T01:16:09Z

Flags

← Older revision		Revision as of 01:16, 28 May 2016
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	=Flags=		=Flags=

	{{~~RadioFlag~~}}		{{GeneralFlag}}