Problem 24-2 Source:
Problem elaborated by the author of the site.
In the circuit shown in the Figure 24-02.1, determine:
a) to t = 0+ the values of i1, i2,
i3, i4, eC, eL.
b) repeat calculation of item variables a) to t → ∞
Figure 24-02.1
Solution of the Problem 24-2
Item a
To solve this problem, let us remember that a capacitor acts as a short circuit when it experiences a sudden change in voltage between its terminals. And, in turn, the inductor behaves like an open circuit in the same situation. So one can represent this state in
t = 0+, in the circuit as shown in the Figure 24-02.2.
Figure 24-02.2
To calculate the current i1, we must calculate the
equivalent resistance of the circuit. To do so, the two resistors of
2 Ω which are in parallel can be replaced by a resistor of
1 Ω. And this resistor is in series with the resistor of 1 Ω, totaling Req = 2 Ω.
Applying the Ohm's Law, we find i1, or:
i1 (0+) = V / Req = 20 / 2 = 10 A
Notice from the circuit that i4(0+) = 0, because the inductor is an open circuit. To find the values of i2 and i3, a current divider may be used. However, in this case, both resistors have the same ohmic value. Soon, the current will split into two equal parts. Thus we have:
i2 (0+) = i3 (0+) = i1 (0+) / 2 = 5 A
Note that since the capacitor is shorted, the voltage between its terminals is equal to zero . In the case of the inductor, the voltage between its terminals is determined by the voltage drop across the resistor 2 Ω. So:
eC (0+) = 0 e eL(0+) = 2 x 5 = 10 V
Item b
When t → ∞ the capacitor behaves like an open circuit and the
inductor behaves like a short circuit. The circuit shown in the Figure 24-02.3
details this situation.
Figure 24-02.3
Note that the inductor short circuits the two resistors of 2 Ω and the capacitor. Hence we conclude that:
i2 (∞) = i3 (∞) = 0 A
On the other hand, we note that i1 = i4. So:
i1 (∞) = i4 (∞) = 20 / 1 = 20 A
Obviously, from the circuit, it is concluded that:
