# Fundamental Electrical Circuits Practice Problem on First Order Circuits

#### Kunchala Anil

Joined Jul 23, 2015
2

#### WBahn

Joined Mar 31, 2012
26,398
The first step is to show YOUR best attempt to solve this.

#### Kunchala Anil

Joined Jul 23, 2015
2
Hi,

when switch is opened for long time, the voltage across capacitor is 15v
when the switch is closed at t = 0 , it become an initial voltage so v0 = 15v
in order to find v_inf across capacitor after the switching takes place.. the open circuit voltage across capacitor by applying kvl across loop
7.5+6i1+2i1+15 = 0
which gives
i1 = 2.81

so the voltage across Open circuit
15-2*i1 + -(7.7-6*i1) = 9.376

the eq for Rc transient response is
v_inf + (v_0 - v_int) exp(-t/tau)

9.376 + (15 - 9.376) exp(-t/tau)
9.376 + (5.624)exp(-t/tau)

the value of time constant tau = R * C
r = 2||6 = 1.5
tau = 1.5 * 1/3 = 0.5 sec.

It did take some time to get ie.. But i think i solved it..

#### WBahn

Joined Mar 31, 2012
26,398
Hi,

when switch is opened for long time, the voltage across capacitor is 15v
when the switch is closed at t = 0 , it become an initial voltage so v0 = 15v
in order to find v_inf across capacitor after the switching takes place.. the open circuit voltage across capacitor by applying kvl across loop
7.5+6i1+2i1+15 = 0
which gives
i1 = 2.81
You need to define what direction i1 is in. Don't make people (e.g., a grader) guess or reverse engineer your work.

You appear to be summing voltage gains going clockwise around the loop (or, equivalently, voltage drops going counterclockwise). Thus you appear to have defined i1 to be the current going counter-clockwise.

You also need to start tracking your units properly. 7.5 is not a voltage, it is just a number. 6i1 is not a voltage, it is a current. 2.81 is not a current, it is just a number.

7.5 V + 6 Ω·i1 + 2 Ω·i1 + 15 V = 0

Look at the circuit and ask yourself what direction current is flowing when the switch is closed (in steady state). It is flowing clockwise, meaning that i1 has to be negative.

Then consider the equation you have. If i1 turns out to be positive, then every term in the above equation is positive, so how can the sum of them possibly be zero.

Thus you have TWO things screaming at you that i1 HAS to be negative. If you come up with a positive value, you KNOW you have made a mistake.

i1 = - (7.5 V + 15 V) / (6 Ω + 2 Ω) = -2.812 A

Answers are generally given to three sig figs, so you want to carry at least one or two more in the intermediate results to minimize roundoff errors.

so the voltage across Open circuit
15-2*i1 + -(7.7-6*i1) = 9.376
Huh? This equals 18.54 (no units since you don't believe in units).

The voltage across the capacitor is

15 V + 2 Ω·i1 = 15 V + (2 Ω)(-2.812 A) = 9.376 V

You are now using i1 as if it is defined to be flowing clockwise, which is inconsistent with your KVL equation. But, in doing so, you have added a second error that, by coincidence, cancels out the sign error you made earlier. Were I grading this, you would have gotten very low marks because you made two mistakes and failed to catch either because of sloppiness. The fact that they happen to cancel out buys you nothing.

the eq for Rc transient response is
v_inf + (v_0 - v_int) exp(-t/tau)

9.376 + (15 - 9.376) exp(-t/tau)
9.376 + (5.624)exp(-t/tau)

the value of time constant tau = R * C
r = 2||6 = 1.5
tau = 1.5 * 1/3 = 0.5 sec.

It did take some time to get ie.. But i think i solved it..