5. For the circuit depicted in the following figure, determine the voltage labeled v₁ across the 3 resistor. +½- 302 292 5 Ω www www 2 A 4VI + VI
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- Using the rules for parallel circuits and Ohmslaw, solve for the missing values. ETE1E2E3E4ITl1I2l33.2AI4RT3.582R116R210R3R420PTP1P2P3P4What is the value of Rin ohms such that the current through the 25 ohm resistance is zero amperes. 100 45 0 25 Q 100 V RO 20 Q Type your numencarswen pnc sumN 7:52 Et C S/10/2 99+ +12 R1=100ohm R2=50 ohm R3=50ohm E=6V R4=75ohm I. Find 15 and 1
- The voltage divider shown in the figure has a potential difference (V) between point A and B and a potential difference (VO) between points C and D. What is V FO? A B RM 2R O A. 2 OB. 4 O C.3 OD. S M 2R R :2R C DWrite down the steps on how to Series Resistors Voltage Divisiona) Voltage at Node "A" in V is b) Current flowing through R5 in mA is c) Current flowing through R6 in mA is d) Current flowing through R7 in mA is e) Power dissipated by R8 in micro W is
- 1. On a single V-I plot, plot the current through the following resistors as a function of voltage. Let voltage range from -10V to 10V and clearly label each axis. R=1000, 2500, and 5000C -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…In 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 R6