The switch in the above circuit has been closed for a long time before it is opened at t=0. Find:
- iL(t) for t>0-;
- i0(t) for t>0–
- v0(t) for t>0–
- The percentage of the total energy stored in the 2 H inductor that is dissipated in the 10 Ω resistor .
The switch has been closed for a long time prior to t=0, so the voltage across the inductor must be zero at t=0–. Therefore the initial current in the inductor is 20 A at t=0–. Hence, iL(0+) also is 20 A because an instantaneous change in the current cannot occur in an inductor. Replacing the resistive circuit connected to the terminals of the inductor with a single resistor of 10 Ω:
The initial energy stored in the 2 H inductor is
Therefore the percentage of energy dissipated in the 10 Ω resistor is
Percentage dissipated = 256/400 (100)= 64%.
The switch has been in position (a) for a long time. The switch then moves to the other position (b).
- Find the expression for i(t) at the position (b),
- What is the initial voltage across the inductor just after the switch has been moved to position (b),
- Does this initial voltage make sense in terms of circuit behavior?
- How milliseconds after the switch has been moved does the inductor voltage equal 24 V?
- Plot v(t) versus t .
- The switch has been for a long time in position (a), so the inductor is a short circuit across the 8A current source. Therefore the inductor carries an initial current of 8 A. This current is oriented opposite to the reference direction for i; thus I0 is –8 A. When the switch is in position (b), the final value of i will be 24/2=12 A. The time constant of the circuit is 200/2 or 100 ms. Then :
- Yes, in the instant after the switch has been moved to position (b), the inductor sustains a current 8 A which causes a 16 V drop across the 2 Ω resistor. This voltage drop adds to the drop across the source, producing a 40 V drop across the inductor.
- Solving the expression:
then t=51.08 ms.
The switch in the circuit has been in position x for a long time. At t=0, the switch moves instantaneously to position y. Find:
- vc(t) for ,
- vo(t) for ,
- io(t) for and
- The total energy dissipated in the 60 KΩ resistor.
The switch has been in position (1) for a long time. At t=0, the switch moves to position (2). Find:
- vo(t) for and
- io(t) for .
- The initial value of vo is 40(60/80) = 30 V. Find the Norton equivalent with respect to the terminals of the capacitor for , by first computing the open-circuit voltage, which is given by –75 V source divided across the 40 KΩ and 160 KΩ resistors: