The initial voltage across the capacitor shown in Figure P4.3 is
Figure P4.3
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- When we talk about capacitance of capacitor we normally say that capacitance depends on the size, shape, and position of the two capacitors and dielectric constant K. What then did we mean when we say that capacitance is constant in the equation Q = CV?arrow_forwardFor the circuit, now keep switch J2 closed, close switch J1. Determine the expressions for the voltage across the capacitor and the current delivered by the source. Find the time it takes for the capacitor to reach its final voltage and what value it is.arrow_forward4.4 25 25. A capacitor (0.02 F) is charged to 1 V and then connected in series with an inductor (10 H) and a resistor (40 S2). Ini- tially, there is no current in the circuit. Find the amplitude, frequency, and phase of the charge on the capacitor and plot its graph.arrow_forward
- Can you also answer this? Please do the last question. StatedΒ construct a graph of time versus current and voltage across the capacitor. The graph is computerized.Β Will get a like.arrow_forwardHow does the product rule of differentiation (differential calculus) useful in Capacitance and Capacitors in Series and Parallel Connections?arrow_forwardThe capacitor behaves as an open circuit to a DC source. Why? How does the inductor behave to a DC source? Please explain in great detail.arrow_forward
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- A capacitor can be used .............................................. in electrical circuits. Which of the followings fit? Β To regulate the currents To regulate the voltages To store energy by deposition of electric charges To provide energy as passive circuit element To store energy in the form of magnetic field To create time varying voltages and currents To dissipate energy To convert elektrik energy into heat or lightarrow_forwardSuppose that we have a 1000-pF parallel-plate capacitor with air dielectric charged to 1000 V. The capacitor terminals are open circuited. Find the stored energy. If the plates are moved farther apart so that d is doubled, determine the new voltage on the capacitor and the new stored energy. Where did the extra energy come from?arrow_forwardAt 0-, no currrent flows through the capacitors because they are open, how did you combine the capacitors for the voltage divider since it is the capacitance value and not reactance, is it right? Can we just combine the capacitance value? Please explain. I did not understand..arrow_forward
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