سيتم دراسة تصرف دائرة مقاومة مع مكثف في حالة التغير الفجائي للجهد أو التيار . و الشكل التالي يوضح الدائرة المستخدمة للدراسة

و باستخدام مكافئ نورتن ، و بحساب مجموع التيار في النقطة العلوية نحصل علي العلاقات التالية :

Example1

The switch in the above circuit has been closed for a long time before it is opened at t=0. Find:

1. i_L(t) for t>0^-;
2. i₀(t) for t>0^–
3. v₀(t) for t>0^–
4. The percentage of the total energy stored in the 2 H inductor that is dissipated in the 10 Ω resistor .

Solution:

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, i_L(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%.

Example2

The switch has been in position (a) for a long time. The switch then moves to the other position (b).

1. Find the expression for i(t) at the position (b),
2. What is the initial voltage across the inductor just after the switch has been moved to position (b),
3. Does this initial voltage make sense in terms of circuit behavior?
4. How milliseconds after the switch has been moved does the inductor voltage equal 24 V?
5. Plot v(t) versus t .
   

Solution:

1. 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 :
   
   
   1. 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.
   2. Solving the expression:

          then t=51.08 ms.

   
   

Example3

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:

1. v_c(t) for ,
2. v_o(t) for ,
3. i_o(t) for and
4. The total energy dissipated in the 60 KΩ resistor.

Example5

The switch has been in position (1) for a long time. At t=0, the switch moves to position (2). Find:

1. v_o(t) for and
2. i_o(t) for .
   

Solution:

1. The initial value of v_o 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: