Concept explainers
The current
Answer to Problem 5.65HP
The value of current
Explanation of Solution
Calculation:
The given diagram is shown in Figure 1
The conversion from
The conversion from
The conversion from
The conversion from
The conversion from
The conversion from
The conversion from
For time
The required diagram is shown in Figure 2
From above, the expression for the current
The inductor opposes the sudden change in the current, thus the current
Substitute
The expression for the voltage across the capacitor for time
Substitute
The expression for the voltage across the capacitor for time
Substitute
The capacitor opposes the sudden change in the voltage and acts as a short circuit. Change the switch position, mark the values and redraw the circuit for
The required diagram is shown in Figure 3
Apply KCL to the top node of the above circuit.
Apply KVL in the left loop of the above circuit.
Substitute
Substitute
The roots of the above differential equation are given by,
The expression for the voltage
Substitute
Substitute
Substitute
The differentiation of
Substitute
Substitute
Substitute
Substitute
From above and from equation (3), the evaluated value of the constant
Substitute
Substitute
Substitute
Substitute
Substitute
Conclusion:
Therefore, the value of current
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Chapter 5 Solutions
EBK PRINCIPLES AND APPLICATIONS OF ELEC
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- A solid specimen of dielectric dielectric constant of 4.0, shown in the figure has an internal void of thickness 1mm. The specimen is 1cm thick and is subjected to a voltage of 80 kV(rms). If the void is filled with air and if the breakdown strength of air can be taken as 30 kV(peak)/cm, find the voltage at which an internal dischargearrow_forwardWrite the differential equation for t > 0 for thecircuit of Figure P5.34arrow_forwardFind the phasors for the current and the voltages for the circuit shown in Figure P5.45. Construct a phasor diagram showing Vs, I, VR, and VC. What is the phase relationship between Vs and Iarrow_forward
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