A 1kV battery, a 3 kW resistor, and a 0.50-mF capacitor are connected in series with a switch. What is the current in the circuit at a time interval equal to twice the time constant after the switch has been closed? 0 A 45 mA 3 A 0.33 A
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- A 1kV battery, a 3 kW resistor, and a 0.50-mF capacitor are connected in series with a switch. What is the current in the circuit at a time interval equal to twice the time constant after the switch has been closed?
- 0 A
- 45 mA
- 3 A
- 0.33 A
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- In the following circuit, V1 = 80 V and V2 = 98 V and all of the resistors are 18 Ω Find the current labelled I2In the circuit shown, how much voltage is used by D? A E 5.58 V 5.06 V 6.11 V 4.38 V 9V 4500 3500 3000 4002 2502A 6.00 nF capacitor that is initially uncharged is connected in series with a 900 2 resistor and an emf source with & = 360V and negligible internal resistance. A long time after the circuit is completed (enough time that capacitor is fully charged), what is the current across the resistor? O 0.20 A O 0.067 A 1.33 A OA 0.24 A O 0.40 A 0.16 A
- Consider the circuit shown below. The resistor has R= 40 ohms and the capacitor has C= 6 x10^-6 F. Initially the switch is open and there is no charge on the capacitor and no current in the resistor. At a time after the switch is closed the current in the resistor is 2 A and the charge on the capacitor is q= +3.00 x 10^-4 C. What is the emf of the battery?An initially uncharged 3.31 × 10-6 F capacitor and a 8430 № resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current Io in the circuit? Calculate the circuit's time constant t. How much time t must elapse from the closing of the circuit for the current to decrease to 3.53% of its initial value? Io = T = t = A S SA capacitor charging circuit consists of a battery, an uncharged 20 μF capacitor, and a 4.0 kΩ resistor. At t = 0 s, the switch is closed; 0.15 s later, the current is 0.46 mA. What is the battery’s emf?
- The following equations describe an electric circuit. −I1(247 Ω) + 5.80 V − I2(381 Ω) = 0 I2(381 Ω) + I3(150 Ω) − 3.10 V = 0 I1 + I3 − I2 = 0 Calculate the unknowns (in mA). With respect to the 5.8 V battery, consider current moving toward the positive pole as positive and current moving toward the negative pole as negative. THE FOLLOWING GIVEN ANSWERS WERE INCORRECT: I1= -10.008 mA I2= -0.868 mA I3= 0.139 mATwo resistors have resistances R1 and R2, where R1 = 2.8R2. When configured in series and connected to a source potential of 23.6 V, they draw 0.723 W of power. If their configuration is changed to be in parallel and the source potential remains the same, what power does the circuit draw