= = (600.2) (1110.8889 μA²) 225 V2 + 300 Ω = 0.666 W + 0.75 W Total power absorbed = 1.416 W

8.

9 kHz R₁ 2200 w R₂ 1800 1 mH .C 0.47 µF

What does s/c and 115 ohms mean?

The resistance of a 1 kW electric heater when energized by a 230 V 1-phase AC is A 52.90 B 230 C 1000Ω D 4.20

For the given figure, v = 72/20°V. Find the average power absorber by the 100 resistor. 492 -j592 0.1V 810 j5Q + 1092 Vo The average power absorber by the 100 resistor is kW (Calculate to 2 decimal places)

Which of the following circuits would yield the same measurements as the circuit shown in the video?

Consider the following table obtained during Lab 1. Instrument Ruler Quantities Diameter Units 191 Meas #1 Meas #2 Meas #3 Meas #4 Meas #5 Meas #6 0.011 0.017 0.019 0.016 0.009 0.021 Vernier Caliper Diameter 0.0120 0.0166 0.0182 0.0151 0.0100 0.0225 Micro-meter Diameter 0.01268 0.01745 0.01904 0.01587 0.00947 0.02221 Ruler Volume 7.0E-07 2.6E-06 3.6E-06 2.1E-06 4.E-07 4.8E-06 Vernier Caliper Volume 9.05E-07 2.40E-06 3.16E-06 1.80E-06 5.24E-07 5.96E-06 Micro-meter Volume 1.067E-06 2.782E-06 3.614E-06 2.093E-06 4.45E-07 5.736E-06 Scale Mass kg 0.00850 0.02080 0.02720 0.01680 0.00389 0.04423 Using Microsoft Excel, or any other graphing software, plot the mass of all six spheres along the y-axis and their respective volumes for the Vernier caliper along the x-axis, obtain the determination coefficient, R², and write your answer below.

Consider the circuit below, what is the current flowing at 8 kohms? 24 1 ev. 0144 mA 0096 mA 0/2.91mA 058 mA

Find the following: (Please show complete solution) - Current after 3.3kohms resistor - Current after 1Okohms resistor - Current after 50kohms@50% variable resistor Given ans.: 720.825 µA, 512.396 µµA, 208.43 µA AIVIIVI II 720.825µA S9 R17 3.3k R18 10k 50% V9 www RP2 50k. RMM12 XMM13 LED LED1 1.497V 1.411V