Concept explainers
Assume the switch in the circuit in Figure P5.69has been closed for a very long lime. It is suddenly opened at
The inductor current
Answer to Problem 5.69HP
Explanation of Solution
Given:
The given circuit is shown below:
Calculation:
When switch is closed for long time means the circuit is in steady state condition. In this condition the capacitor is open circuited, and inductor is short circuited. Therefore, the modified circuit can be drawn as:
The current through the source just before the opening it at t = 0,
Applying the current division rule, the initial value of the current can be calculated as
The initial value of voltage across the capacitor will be
The switch is opened at t = 0, the circuit can be drawn as
Applying KVL in the loop having current
Applying KVL in the loop having current
Differentiating with respect to t,
Put the value of
Assuming
The auxiliary equation can be written as
The roots can be calculated as
Therefore, the solution of the differential equation will be in the form
Current through the inductor is equal to the loop current,
Using the initial conditions,
At t = 0,
Substituting 0 for t,
Therefore,
Differentiating equation 2 with respect to t,
Substituting 0 for t,
Applying KVL in the loop with current
Substituting the values,
Substituting this in equation 4,
Substituting A and B in equation 2,
Substituting A and B in equation 3,
Put these values in equation 5,
Substitute
Substitute
Reclose the circuit. Now, the initial current in the inductor is represented as a voltage source of
Thus,
The initial voltage across the capacitor is represented as a voltage source of
The circuit can be drawn as
It can be observed that the voltage across the
Apply KVL in the loop with current
Apply KVL in the loop with current
Assuming
The auxiliary equation can be written as
The roots can be calculated as
Therefore, the solution of the differential equation will be in the form
Current through the inductor is equal to the loop current,
Using the initial conditions,
At t = 5,
Substituting 5 for t,
Differentiating equation 8 with respect to t,
Substituting 5 for t,
The voltage across the inductor at t = 5s will be
Substituting this in equation 10,
Substituting A and B in equation 8,
The complete solution can be calculated as
The 6V is always connected across the 2ohm resistor whenever the switch is closed. Therefore, the voltage across the 2ohm resistor will be
When the switch is opened, the current through the inductor flows through the 2ohm resistor. Thus, the voltage across the 2ohm resistor, when the switch is open will be
Substituting the value of
The voltage across the 2ohm resistor will be
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Chapter 5 Solutions
Principles and Applications of Electrical Engineering
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