R1 R3 R5 V1= IA E R2 IB R4 Ic EV3 V2 Calcuate the values of the indicated mesh currents IA= A,IB= A, and IC= A, if R1=12 Ohms,R23D5 Ohms,R3=24 Ohms,R4%3D22 Ohms,R5=D9 Ohms,V136 Volts, V2=5 Volts,and V3=10 Volts.
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- Estimate voltage E2 when E1=90sin2400t V, R1=R2=20 ohms, L1= L2=3mH and M=0.15mHDetermine E and P3. Given R1 = 10ohms, R2 = 20ohms, R3 = 60 ohms, R4= 10ohms, R5= 20ohms, R6= 60ohms, R7= 10 ohms, R8= 20 ohms. R9= 30ohms, P6 = 540W.In the figure: R1 = 27.3 Q, R2 = 93.1 Q and R3 = 1571.12 Q, then determine the value of the source current Is. 12 13 RT R1 R2 R3 E 1.33 0.98 1.44 1.73 1.15
- Determine the DC currents (IB, Ic and lĘ) and dc junction voltages (VBE, VCE and VCB) Ig=Blank 1 mA, Ic=Blank 2 mA, IĘ=Blank 3 mA, VBE= Blank 4 V, VCE = Blank 5 V and VCB = Blank 6 V Use 2 decimal places. Use the following values: VBB = 2V RB = 7 kQ Rc = 248 Q Vcc = 19 V BDc = 56 Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answer Blank 4 Add your answer Blank 5 Add your answer Blank 6 Add your answerIn the figure the current in resistance 6 is i6 = 1.33 A and the resistances are R₁ = R₂ = R3 = 1.56 02, R4 = 14.7 Q, R5 = 7.74 Q2, and R6 = 3.81 Q. What is the emf of the ideal battery? 80 R₁ Number i www R₂ www R₂ Units www R₁ www R₂ i6 www R6The figure below shows three resistors (R- 13.5 0, R -7.95 0, and Ry- 12.5 0) and two batteries connected in a circuit. 40.0 V R 22.0 V R3 (a) What is the current in each of the resistors? A I2 = A A (b) How much power is delivered to each of the resistors? P1- P2- P3-
- V1= 1.5Vkn=0.25mA/V2Find Q-pointTwo resistors RA and RB are connected in series across a 120-V source. When a 30,000-ohm voltmeter is connected across RA and then across RB, the instrument deflection is 48 volts in each case. What are the ohmic values of the resistors?C -Q (A) 0.25 uF 10V 2uF 4uF (B) Figure 3: 3. In Figure 3 the capacitors were discharged before being connected to the battery. (a) In Figure 3A the net positive charge delivered by the battery is +Q1 to C1 and a net negative charge -[Q2 + Q3] is delivered to C2 and C3 [-Q2 is delivered to C2 and -Q3 to C3]. From conservation of charge[net charge is zero since both the battery and the capacitors have a zero net excess charge] and path independence of the electric field[sum of voltages must add up to zero for a closed path or conservation of energy]: Q1 + (-Q2) + (-Q3) = 0 conservation of charge +%-왕-용 (True, False) = 0 conservation of energy +V - = 0 conservation of energy (b) If (a) is true then the charge Q1, Q2 and Q3 follow from above three equations as C1(C2 + C3) V Q2 = Ci + C2 + C3 C1 C3 Ci + C2 + C3 Q1 = Q3 = V, (True, False) C1 + C2 + C3 (c) In Figure 3A the net positive charge delivered by the battery is Qtot = Q1 = C1(C2+C3) v which implies an equivalent capacitance of Qtot…
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