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- The voltage source is at a constant level for a long time, which we approximate as infinite. This is the circuit: The voltage drop across the capacitor rises from 0 to ℰ. Note that ℰ is never actually known in the measurement. In fact, the oscilloscope voltage is decalibrated, so that, whatever ℰ is, ℰ is at the top line while zero is at the bottom line. We don't measure voltage levels, but rather 1/2, 1/4, and 1/8 the maximum. Kirchhoff's voltage law give: ℰ = IR + Q/C or the following: dQdt=−1RC(Q−EC)dQdt=−1RC(Q−ℰC) The solution for the capacitor voltage is VC(t)=E(1−e−t/RC)VC(t)=ℰ(1−e−t/RC) The voltage is zero at t = 0, t is the rising time, and you have to know when the rising begins.In Fig. 6-27e A closes at t = 0 and B closes at t = 5 ms. What is the current waveform through thefirst 2- kΩ R.In the circuit of the image, the capacitor is initially discharged, with the switch open. Values are in the second image.When time t = 0, the switch is closed. Which of these is correct: A. A lot of time has passed after the switch was closed. Now you open it, and set t = 0. When t = 5x10^-3 seconds, the current is 0.036AB. The current given by the battery when t = 5x10^-3(s) is 0.036AC. A lot of time has passed after the switch was closed. The current given by the battery is 0.01 A.D. A lot of time has passed after the switch was closed. Now you open it, and set t = 0. The time it takes the capacitor to decrease to 10% of its max charge, is 0.023 seconds.E. The current when t = 0 is 0.02A
- The current in a 0.6 microfarad capacitor is 0 [A] for time less than zero and 3cos50000t [A] for time greater than or equal to zero. Find v(t) and the maximum power delivered to the capacitor.An LR circuit includes a resistor of resistance R, an inductor of inductance L and a battery of emf E = 10 V. At time t = 0 the current in the circuit is I = 0. At time t = 6.1 ms the current is I = 0.66 A. What are the values of L and R?In an R-L-C series circuit a maximum current of 0.5 A is obtained by varying the value of inductance L. The supply voltage is fixed at 230 V, 50 Hz. When maximum current flows through the circuit, the voltage measured across the capacitor is 350 V. What are the values of the circuit parameters? ..
- How long does it take for an initial charge of 20 V DC to decrease to 1 V DC in a 0.01microfarad capacitor when a 2-megohm resistor is connected across it?A) Reduce the given circuit to the fewest possible componentsthru series/parallel combinationsB) DetermineVx if all resistors are 10k , all capacitors are 50uF and all inductors are 1mHWhich of the following is true when charging a capacitor using a DC source? Current will flow due to the DC source and will charge the capacitor. Once the capacitor is fully charged, it has to be removed from the circuit it will not exceed the voltage of the source to avoid short-circuiting. Since capacitors behave as open circuits when a DC source is applied, there will be no charging current. The charging will happen due to KVL where the voltage of the source will be charged to the capacitor. Current will flow due to the DC source and will charge the capacitor. Once the capacitor is fully charged, it will act as if it is an open circuit. The answer cannot be found on the other choices.
- two capacitors with capacitances of 1.22uF and 6.87uF are connected in parallel. These are connected to a 48V battery . How much energy will be stored in the 6.87uF capacitor in mJ? Please do fast ASAPThe circuit elements in the circuit in (Figure 1) are R = 2k*Omega, L = 250mH and C = 10nF . The initial inductor current is -30mA, and the initial capacitor voltage is 90V.In the definition given below, mark the most appropriate option that will come to the space. When capacitors are examined in terms of DC ................................. circuit is accepted. A-Closed B-Long C-All D-Short E-Open