Calculate the current in the 10-ohm resistor using Kirchhoff's law I, E₁1 = 10 2 10 E₂=4 3
Q: Refer to the Figure below, calculate Z₁ . Assume VGSQ = -2.2V. (VDD = 12V, RD=3.3kohms, Rs=1.3kohms,…
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Q: VGSQ = -2.2V. (VDD = 12V, RD=4.1kohms,
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- Using the rules for parallel circuits and Ohmslaw, solve for the missing values. ETE1E2E3E4ITl1I2l33.2AI4RT3.582R116R210R3R420PTP1P2P3P4Here are the choices: Lorentz law charge Faraday's law Ohm's law Capacitance Resistivity of materials Resistance vs Strain2Q -Q 2r (a) -Q -2Q (b) D- D- 2r 5- r (c) (d) What is correct order of the charge systems fro c>a>d>b A)
- What 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+ +If a circuit has a resistance of 3002 and a series inductance of 5002, what is the relationship between the applied circuit voltage and the circuit current? The applied voltage lags the circuit current. The phase shift between the circuit voltage and current changes with time. The applied voltage is in phase with the circuit current. The applied voltage leads the circuit cur Previous Page Next Page Page 2 of 10determine the currents and the voltages with kirchhoff's law 2200 A 0.3 150N A 4V C 200N A
- The conductance of a 9230 resistance is a) 1.08 * 10-3 mho b) 1.08 * 10-4 mho c) 1.02 * 10-3 mho d) 1.02 * 104 mhoDetermine the loop currents Ia, I,, and Ic. 100 Ic 100 100 100 100 Ib 55V 100 V laHow much resistance must be inserted in series with a 0.96-H inductance to limit the current to 0.25 A from a 120-V 60 Hz power line?
- Three 47.1 ohm resistors and three 17.1 H inductors are all in series with a 37 volt battery and a switch. The switch is placed in position A at time t = 0 and switched to position B at time t = 743 ms. Find the voltage across each of the resistors at time t = 1.17 seconds. Owo DO A R R in in ete SW éte لفظوC -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…For some reason it's saying the 0.3636 is wrong. I tried 0.364 as well thinking it may be have wanted it rounded. This is explained very well. Thank you. On the second part, you did 60-60 because the charge didn't change. Is that correct?