Derive an expression for V(t) for the circuit below. Solve for kj and k2 in V(t) equation using initial conditions.
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- The figure below shows a simple RC circuit with a 3.30-μF capacitor, a 4.40-M resistor, a 9.00-V emf, and a switch. What are the following exactly 9.00 s after the switch is closed? (a) the charge on the capacitor 13.7204 UC (b) the current in the resistor 0.0110522 x Your response is off by a multiple of ten. HA (c) the rate at which the capacitor is storing energy μW (d) the rate at which the battery is delivering energy μWFor the circuit below with input voltage V1, a step function of magnitude 25 volts at time t-0, find the transient response voltage across the capacitor C1 and plot the results: (Provide your calculations and reasoning for your answer.) R1 555 250mH V1 = 0 V2 - 25 TD = 0 TR In V1 C1 TF in PW- 25 PER = 3.3uFor R1=9000, R2=2000, R3=6000, R4=5000, C1=0.006, C2=D0.002 & 11=0.007 A in the shown circuit, obtain the energy stored in each capacitor under dc conditions. C2 R4 ŽR3 ER2 ŽRI C1= Energy stored in C1= Energy stored in C2=
- 6. PLEASE SOLVE AND SHOW YOUR DETAILED SOLUTION WITH FBD. The currents at the junction point in a circuit have the following values I1 = (15 – j4) A and I2 = (12 + j22) A. Calculate I1 + I2.Initially relaxed series RLC circuit with R = 150 £2, L=30 H and C= 10 F has de voltage of 150 V applied at time t = 0. Find the equation for current and voltages across different elements. Also draw the waveform under different cases. E R i L -00000 CIn the circuit shown in figure the switch is closed at time (t=0). The rate of change of current dt across inductor will be tm0 20 8F 20v (3H
- A series R–L–C circuit comprises a 5µF capacitor, a 4ohm resistor and a variable inductance L. The supply voltage is 10∠0◦ V at a frequency of 159.1 Hz. The inductance is adjusted until the p.d. across the 4 ohm resistanceis a maximum. Determine for this condition (a) the value of inductance, (b) the p.d. across each component and (c) the Q-factor of the circuit.The figure below shows a simple RC circuit with a 3.50-µF capacitor, a 3.60-MQ resistor, a 9.00-V emf, and a switch. What are the following exactly 6.50 s after the switch is closed? S C R (a) the charge on the capacitor (b) the current in the resistor µA (c) the rate at which the capacitor is storing energy (d) the rate at which the battery is delivering energyA capacitor of capacity C₁ is charged upto V volt and then connected in parallel to an uncharged capacitor of capacity C₂. The final potential difference across each will be (a) (c) C₂V G₁ + C₂ (1+2) GV G + C₂ (4) (1-²) v (b)
- The circuit in Figure has been connected for a long time. If the battery is disconnected, how does it take the capacitor to discharge to 1/20 of its initial voltage? Take R₁ = 2.00 Q, R₂=4.00 Q, R3-7.00 Q, R4 = 2.00 Q and C= 5.00 uF and ε = 14.0 V. (Your result must be in us. Include 2 digit after the decimal point and maximum of 2% of error is accepted in your answer.) E R₂ R₂ BAFor the circuit shown in figure Req is: * 140 R3 R, R: 22 Rea Ru RC RS R7 33 6 ohm 24.4 ohm 34,4 ohm 44.4 ohmThe capacitor in the circuit is initially uncharged and the switch S is suddenly closed at t=0. Find the current on resistor R: at to Take R₁-4R and R 3R. A) = B) 6 25 D) E) SIA R₂ FF R₁