Q: y4 60° A = 10.0 30° 30° B 53° D D = 20.0 37° C = 12.0 10 F = 20.0 B = 5.0
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Q: 60° A = 10.0 30° 30° 53° D 37° C = 12.0 F = 20.0 B = 5.0 D = 20.0
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Q: R1 = 5A I tot =. Rgot AV1=. R2= R3= 12A AV2=- 12=. 24V AV3-. I3=. AV4=- 14=. R4=8A
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Q: B = 6 [u] C = 8,96 [u] A = 12,75 [u] a = 23.98° B = 28.1 y = 40.04 O = 21.1° E = 11,71 [u] D = 6,86…
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Q: 5.0 μΕ. 15 μΕ 8.0 µF 3.5 µF 0.75 uF 15 µF C1 = 5.0 µF C2 = 3.5 µF C3 = 8.0 µF C4 = 1.5 µF Cs = 0.75…
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By using mesh analysis
At loop (1)
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- Reclung.c A on ri gle Figure 1 2. The circuit shown above is constructed with two 6.0 V batteries and three resistors with the values shown. The currents 1₁, 12, and 13 in each branch of the circuit are indicated. 2 GEOMETRY AND TRIGONOMETRY Circle A = ²² C = 2πr Cy.inder bh - re Rectangular Solid Vlh Sphere 1. Using Kirchhoff's rules, write, but DO NOT SOLVE, equations that can be used to solve for the current in each resistor. ii. Calculate the current in the 200 9 resistor. iii. Calculate the power dissipated by the 200 2 resistor. V = ar²l S.. Figure 2 The two 6.0 V batteries are replaced with a battery with voltage & and a resistor of resistance 50 2, as shown above. The voltmeter V shows that the voltage across the 200 £2 resistor is 4.4 V. (b) Calculate the current through the 50 2 resistor. (c) Calculate the voltage & of the battery. V=² S = 4; 7² Rig Triangle 2 +6² = c² sin = COS== tan 8 = OJA b C 2m. a b 6.0 V 7,4 & A - area C=circumference h-hei £= le L W = IUL r=fuu.0…FET. Answer the following problems. Use the MATHEMATICAL APPROACH unless otherwise specified. Show your COMPLETE SOLUTION. For the given fixed-bias configuration a. Sketch the transfer characteristics of the device. b. Superimpose the network equation on the same graph. c. Determine Ipq and Vpsq: (graphical approach) d. Using Shockley's equation, solve for Ipo and then find VpsQ. Compare with the solutions of part (c). 14 V 1.8 kn Ipss = 12 mA Vpsg Vp = -4 V 1 M2 1.5 V-1. 9. In the circuit shown in the figure above, a power supplier of 50 V is connectedacross a resistance network. If R1 = 20 Ω, R2 = 10 Ω, R3 = 10 Ω, and R4 = 40 Ω. Whatis the equivalent resistor of the circuit? 2. 10. What is the current flowing through the R4 resistor in the circuit shown in Problem 9? 3. 11. What is the voltage across the R1 resistor in the circuit shown in Problem 9? 3.
- Capacitance, Charge and Voltage: Q = CV 1. Click the Reset Button on the bottom right of the PHET simulation. 2. CHECK Plate Charges and Bar Graphs boxes in the upper right so your display resembles the graph below. 3. SET the Plate Area to 200 mm² and the Separation to 4.0 mm. Capactance Tup Pate Charge oc Sured Energy Pu Charges r Graphs O e Fd Cument Direction 1500 V 4. Connect the voltmeter across the capacitor by placing the red on the top plate and black on bottom. If the voltmeter reads a negative, switch the red and black. 5. Set the battery Voltage to 0.25 V. Record the measured plate charge, Q. 6. Repeat Step 5 for each Voltage in Table 3. Table 3: Charge and Voltage Voltage |(V) Plate Charge |(pC) 0.25 V 0.50 V 0.75 V 1.00 V 1.25 V 1.50 VSolve for the given problems. Show your complete solutions. Round off your answers into 3 significant figures Refer to figure 7. a. Find the current in the circuit. b. Find the terminal voltage Vab of the 16.0-V battery. c. Compute for the potential difference Vac.Activity 2 Directions: Solve for the given problems. Show your complete solutions. Express your final answers into 3 significant figures. 1. Silver is composed of 5.8 x 1028 free electrons per cubic meter. If a silver wire with a diameter of 1.8 mm can allow charges of 360 C to flow in 1 hour, a. What is the current in the conducting wire? b. What is the magnitude of the drift velocity of the electrons in the conducting wire? 2. A wire has a radius of 1.90 mm. If the current in the wire is 6.14 A, the drift velocity is 4.86 x 10-5 m/s. What is the concentration of free electrons in the metal?
- What to do: Answer the following problems. Write your solutions in a separate sheet of paper. 1. Compute the equivalent resistance of the network in the circuit shown below. Show the steps in reducing a combination of resistors to a single equivalent resistor. The source of emf has an negligible internal resistance. E = 18 V, r = 01. Does the resistance of all substances increase with temperature? Explain. 2. What is the temperature coefficient of resistance, and what are its units? 3. Are the a of a metal conductor and the B of a thermistor the same? Explain. 4. What the circuit conditions when Wheatstone bridge is are a "balanced"? 5. Assuming that B remained constant, what would be the resistance of the thermistor in the experiment as the temperature approached absolute zero? 6. What characteristics of the thermistor make it flexible a temperature transducer?Imagine building a circuit using a perfect insulator (a device of infinite resistance, which allows no current flow) in place of the carbon resistor a. Would the plot of the data be a straight line? b. If yes, what would the slope of that line be? If no, what shape would the line be?
- The output power of a solar PV module is 120 Watts. The number of cells in the moduleis 39. Estimate the following: (Assume any other necessary data)i. Output voltage of the PV moduleii. Output current of the PV moduleiii. Total area of PV moduleiv. Area of one PV cellv. Power output of one PV cellQ2. Describe the functional difference between a NAND gate and a negative-OR gate. Do they both have the same truth table?Answers to the following question will be relevant for the in-labs. Please note your answers when going in for the main lab session. 3. Ohm's law can be sated as V = IR. a. Suppose in an experiment, data is collected of the variation of current with resistance R. Next a plot of R vs I'is made. How does one determine the value of V from that graph? (Explain your answer using an appropriate expression.) b. Instead of measuring resistance suppose that in (a) data of the variation of current with the length of wire is collected. Show how such data can be used to determine the resistivity of the wire.