ELECTRICITY FOR TRADES (LL W/ACCESS)
ELECTRICITY FOR TRADES (LL W/ACCESS)
3rd Edition
ISBN: 9781260699487
Author: Petruzella
Publisher: MCG CUSTOM
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Chapter 7.1, Problem 3RQ

Convert each of the following:

  1. a. 2,500 Ω to kilohms
  2. b. 120 kΩ to ohms
  3. c. 1,500,000 Ω to megohms
  4. d. 2.03 MΩ to ohms
  5. e. 0.000466 A to microamps
  6. f. 0.000466 A to milliamps
  7. g. 378 mV to volts
  8. h. 475 Ω to kilohms
  9. i. 28 μA to amps
  10. j. 5 kΩ + 850 Ω to kilohms
  11. k. 40,000 kV to megavolts
  12. l. 4,600,000 μA to amps
  13. m. 2.2 kΩ to ohms
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dny dn-1y dn-1u dn-24 +a1 + + Any = bi +b₂- + +bnu. dtn dtn-1 dtn-1 dtn-2 a) Let be a root of the characteristic equation 1 sn+a1sn- + +an = : 0. Show that if u(t) = 0, the differential equation has the solution y(t) = e\t. b) Let к be a zero of the polynomial b(s) = b₁s-1+b2sn−2+ Show that if the input is u(t) equation that is identically zero. = .. +bn. ekt, then there is a solution to the differential
dny dn-1y dn-1u dn-24 +a1 + + Any = bi +b₂- + +bnu. dtn dtn-1 dtn-1 dtn-2 a) Let be a root of the characteristic equation 1 sn+a1sn- + +an = : 0. Show that if u(t) = 0, the differential equation has the solution y(t) = e\t. b) Let к be a zero of the polynomial b(s) = b₁s-1+b2sn−2+ Show that if the input is u(t) equation that is identically zero. = .. +bn. ekt, then there is a solution to the differential
For step a), use equations (2) to find the equation for the input impedance equations (2) are V1 = jwL1I1 + jwMI2 and V2 = jwMI1 + jwL2I2 equation for the input impedance: Z1 = V1/I1 = jwL1 + (wM)2/(jwL2 + ZL)
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