Problem + Hard 25-1 Source:
Problem elaborated by the author of the site.
In the circuit shown in the figure below, the switch S has been open for a long time.
In t = 0 the key is closed. Determine the answer of v(t).
Solution of the Problem + Hard 25-1
With the switch S remained open for a long period we can determine the initial conditions of
i and v(t), because the capacitor behaves like an open circuit and the inductor like a short circuit. Therefore, the current i only depends on the resistor values. Like this:
i(0-) = i (0+) = 12 / 24 = 0.5 A
To determine the value of v(t = 0), simply make a voltage divider for the resistor that is parallel to the capacitor. Like this:
vO (0-) = vO (0+) = 12 x 10 / (10 + 14) = 5 volts
So we have already found the initial conditions of the problem. From now on we will consider the
S key closed. In this case, we realize that the voltage source and the 4 ohm resistor are eliminated from the circuit. In this way we can write the loop equation of the circuit as shown below. In addition, the current i(t) can be written as the sum
of the currents that pass through the capacitor and the resistance R2.
eq. 25-a
Then, replacing the second equation in the first and working algebraically on the result, we find the differential equation that governs the behavior of the circuit after closing the switch S. So:
eq. 25-b
Note that the second equation (above) was found by substituting the numerical values of the circuit components in the first equation. Now we can easily write the characteristic equation of the differential equation and calculate the roots, or:
r 2 + 6 r + 10 = 0
In this case, the roots are complex and therefore the circuit has an under-damped response. The roots are:
r1 = -3 + j2 e r2 = -3 - j2
So the equation that reproduces the circuit behavior is given by:
v (t) = e-3t (B1 cos 2 t + B2 sin 2 t)
eq. 25-c
To find the values of B1 and B2 we must use the initial conditions, which were calculated at the beginning of the problem. Making t = 0, the above equation boils down to:
v (0 +) = 5 = B1
With the value of B1, we can find the value of B2 deriving the
eq. 25c. On the other hand, from the second equation shown in eq. 25-a, we can deduce that:
eq. 25-d
So using the known values for t = 0 of i (t) and v (t), we can calculate the value of d v / d t. Then:
d v / d t = (0.5 / 0.1 ) - 5 / (10 x 0.1) = 5 - 5 = 0
Hence, deriving the eq. 25-c, by making t = 0 and equaling zero, we find the value of B2, that is:
d v / d t = - 3 B1 + 2 B2 = 0
So, doing the numerical substitution and making the calculation, we find the value of B2. In short we have:
B1 = 5 and B2 = 7.5
Now knowing the values of B1 and B2 we are able to write the general solution for the studied circuit, or:
