1. Find the base currents of Q1 and Q2. Vcc f14v Rc 6.8 kN R1 33 k2 Q2 B2= 50 B = 100 RE2 1.2 kN R2 10 k RE2.2 kN
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Q: Find the current I2 %3D V = ? R = 1.00 N %3D V = 12.0 V 2 = ? R2 = 6.00 2 R3 = 13.00 Q V3 = ? %3D
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- a 3 Ω w 5Ω h 6V+ L g Determine Vhe. 4 V + f + 7 V 6Ω W C 2V + d 4Ω ePart G: Use Kirchoff's Law to solve for all the voltage drops and branch currents. R1 R4 70 0 100 O ww ww R2 Vs2 15V 225 2 R5 125 Q ww Vs1 20V R3 Vs3 25 Q 6V ww6. chapter27-question31 A conduction current changes over the according to the I = 15t² – 4t + 2 connection, where I is in amperes and t is seconds. How many electrons pass through the cross section of the wire at the interval t = O and t = 1 s? qe = 1.6x10-1ºC. Option1 Option2 Option3 Option4 Option5 6x1019 5x1019 4x1019 3x1019 2x1019 a) Option3 b) Option1 Option5 d) Option2 e) Option4 Boş bırak
- pls solve the r total, v total, i total and p totalQ.6: Given R1=15N, R2=102, R3=222 and R4=152, find the current flowing through resistance R4. R2 R1 T 8.0V Ra R3 Vo S 12.0V V = k U = k. n.n qi9 j Liki rij Ceg = YN Gi - N 112. What is the total Voltage for the circuit shown below? R1 25 Q V2 6 V V1 9 V wwwwww Show your work... What is the current for the circuit shown? Show your work... R2 15Q R3 30 Q
- Consider the circuit shown. The currents are already drawn for you. b R3 C d 3.002 R4 V1 V2 6.002 6.00V 9.00V I2 13 R1 5.002 R2 V3 4.002 12.0V R5 a f e 2.002Find the voltages across R1,R5, R6 and R7. R4 70 Q ww R1 R5 100 Ω 125 2 ww Vs 96V R2 225 Q R7 R6 120 2 500 2 ww R3 25 Q ww ww ww wwThis circuit has 6 distinct currents. Which of these is an appropriate application of the junction rule? Imm O a. 12+16=1₁ O b. 12+16=15+14 OC 12+15=14 O d. 14+16=13 15 In
- Apply the loop rule to loop abcdefghija in the figure below. E, = 24.0 V b d 0.10 2 R 5.0 2 Rs 20 2 E2 = 48.0 V R2 k 12 a e 0.50 2 40 2 0.20 2 R3 78 2 1s E3 E4 = 36.0 V 6.0 V %3D 1 0.05 Ωh O -(1,R2) + E, - (I,r2) – (1,R4) – (I3r2) - E4 – (1gr3) + E4 - (1,R3) = 0 O -(1,R1) + E, - (I,r2) – (1,R,) – (I3r3) – E, - (I,r3) + E - (I,R4) = 0 -(1,R;) + E, - (1,i) - (1,R5) – (I3r4) - E, - (13r3) + Ez - (I3R3) = 0 O -(1,R2) + E, - (I3r3) – (I„R5) – (I34) – E4 - (I3r3) + E4 - (I3R4) = 0 Additional Materials O ReadingIn the double-loop diagram, V = 3_volts and R = 280_N. Find the currents I,,I2, and I3. Only two out of three of the currents are graded. %3D %3D 8 V 75 L 130 2 V I2 13 3 V 55L mit 15. 1 D. -0.02828 A А. -0.02979_A -0.02546 A B. E. -0.03149_A -0.03258 A F. С. -0.02678 A |--/ 16. I2 D. 0.003681 A 0.003458 A 0.003954 A 0.003308 A А. Е. В. 0.003814 A F. С. 0.003135 A |--/For the series-parallel network c) find the magnitude of voltage Vab- R2 R6 5Ω 13 60 V ab R3 R7 a 4 0 Rg' R 10 0 80 V R4 = 40 8Ω R$ 2Ω %3D O a. 14 V O b.9 V Oc. 40 V O d. 80 V