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Refer to Figure P3.44 and use the principle of super position to determine the current i through
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- A nonideal voltage source is modeled in FigureP3.72 as an ideal source in series with a resistance thatmodels the internal losses, that is, dissipates the samepower as the internal losses. In the circuit shown inFigure P3.72, with the load resistor removed so thatthe current is zero (i.e., no load), the terminal voltageof the source is measured and is 20 V. Then, withRL = 2.7 kΩ, the terminal voltage is again measuredand is now 18 V. Determine the internal resistance andthe voltage of the ideal source.arrow_forwardA 10-mH inductor has a parasitic series resistance of R s =1 Ω, as shown in FigureP3.68.a. The current is given by i( t )=0.1 cos( 10 5 t ). Find v R ( t ), v L ( t ), and v(t). In thiscase, for 1-percent accuracy in computing v(t), could the resistance be neglected?b. Repeat if i( t )=0.1 cos( 10t ).arrow_forwardFor the circuit of Figure P3.22 determinea. The most efficient way to solve for the voltageacross R3. Prove your case.b. The voltage across R3.VS1 = VS2 = 110 VR1 = 500 m R2 = 167 mR3 = 700 mR4 = 200 m R5 = 333 marrow_forward
- The circuit shown in Figure P3.35 is a simplifiedDC version of an AC three-phase electrical distributionsystem.VS1 = VS2 = VS3 = 170 VRW1 = RW2 = RW3 = 0.7Ω R1 = 1.9Ω R2 = 2.3Ω R3 = 11 ΩTo prove how cumbersome and inefficient (althoughsometimes necessary) the method is, determine, usingsuperposition, the current through R1.arrow_forwardfind Rt, It, Pt, P1, P2, P3P4,P5,P6arrow_forwardUse mesh current analysis to find the current i inthe circuit of Figure P3.27. Let V = 5.6 V; R1 = 50Ω ;R2 = 1.2 kΩ; R3 = 330 ; gm = 0.2 S; R4 = 440 Ω.arrow_forward
- 5- self inductorQ3-A current source of 3 mA has an internal resistance of 5 MΩ. Over what range of load resistance is the current source stiff? Plot the diagram and circuit.arrow_forwardFind the Thévenin equivalent of the circuitconnected to RL in Figure P3.58, where R1 = 10Ω ,R2 = 20 Ω, Rg = 0.1 Ω, and Rp = 1 Ω.arrow_forwardIn the circuit shown in Figure P3.33, F1 and F2 arefuses. Under normal conditions they are modeled as ashort circuit. However, if excess current flows througha fuse, it “blows” and the fuse becomes an open circuit.VS1 = VS2 = 120 VR1 = R2 = 2 Ω R3 = 8Ω R4 = R5 = 250 mΩIf F1 blows, or opens, determine, using KCL and nodeanalysis, the voltages across R1, R2, R3, and F1.arrow_forward
- Construct the circuit of figure P3-2 using the bipolar junction transistor (BJT). Please typing format solutionarrow_forwardUsing KCL, perform node analysis on the circuitshown in Figure P3.24, and determine the voltageacross R4. Note that one source is a controlled voltagesource! Let VS = 5 V; AV = 70; R1 = 2.2 kΩ;R2 = 1.8 kΩ; R3 = 6.8 kΩ; R4 = 220Ωarrow_forwardUsing mesh current analysis, find the currents I1, I2, and I3 in the circuit of Figure P3.17 (assume polarity according to I2).arrow_forward
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