The step response of a circuit is its response to a step input, u(t). Knowing the step response of a linear circuit is extremely valuable as we can figure out the circuit's stability. In this problem we will analyze the circuit in Figure 1 for the output voltage, vO(t) given an initial condition. Then you will find the output response of two input steps--a current input, iI(t), and a voltage step, vI(t).
Figure 1
The element values are as follows: R1=1kΩ, R2=1kΩ, R3=2kΩ, CA=0.5μF, and CB=1.5μF.
To begin we note that the capacitor bridge composed of four capacitors can be replaced with one capacitor, CTOT. What is the value of CTOT (in microfarads)?
unanswered
First we consider the circuit in Figure 1 when iI(t)=0A, vI(t)=0V, and vO(0)=1V. What are the time constant, τ, of the output voltage and the output voltage, vO(t), at time t=0.3ms?
τ (in milliseconds)
unanswered
vO(t) (in Volts) at time t=0.3ms?
unanswered
For the remainder of this problem we consider the step response where iI(t)=1u(t)mA, vI(t)=1u(t)V, and vO(0)=0V. Remember that u(t) is the unit step function:
sorry, 8.01 and 6.0 anything are not subjects I respond to. Been there, done that. You are all on your own.
Moreover, I can not see the circuit !
to add capacitors in parallel
Ceq = C1 + C2 because C = Q/V
V the same, charges add
to add capacitors in series
q is the same on both
in general V = C Q
here V = V1 + V2 = C1 q + C2 q
V/q = (C1+C2)
so Ceq = q/V = 1/(C1 + C2)
in general q/C
here V = V1 + V2 = q/C1 + q/C2
V/q = (1/C1+1/C2)
so Ceq = q/V = 1/(1/C1 + 1/C2)
= C1 C2 /(C1+C2)
Here we analyze the time response of the circuit in Figure 3 which contains capacitive and resistive elements and a voltage source, VS. The voltage source is switched between two nodes in the circuit, and we observe the voltage response across C1. All answers in this problem are looking for expressions in terms of VS, C1, C2, R1, and R2.
Initially for t<0 the switch connects the voltage source, VS to node 1 and the circuit is allowed to reach steady-state. What is the steady-state voltage, v(t), across C1 for t<0?
At t=0, the switch is repositioned to connect the voltage source to node 2 and the circuit is allowed to reach steady-state again. What is the voltage, v(t), across C1 after the circuit reaches steady-state?
We can use differential equations to write an expression for the voltage, v(t), across C1 for t>0. The homogeneous and particular solutions can be used to express the voltage, v(t)=H∗e−t/τ+P. Give expressions for H, P and τ.
τ
P
H
@Damon plz help me for this question if u got it correct ??
To find the value of CTOT (in microfarads), we can first simplify the circuit by combining the four capacitors into one equivalent capacitor. The equivalent capacitance can be found using the formula:
1/CTOT = 1/CA + 1/CB + 1/CC + 1/CD
Substituting the given values for CA, CB, CC, and CD:
1/CTOT = 1/0.5μF + 1/1.5μF + 1/1.5μF + 1/0.5μF
Simplifying the equation:
1/CTOT = 2/0.5μF + 2/1.5μF
1/CTOT = 4/0.5μF + 2/0.5μF
1/CTOT = 6/0.5μF
1/CTOT = 12/1μF
CTOT = 1μF / 12
CTOT = 0.0833μF ≈ 83.3nF
Therefore, the value of CTOT is approximately 83.3nF.
Now, let's move on to the second part of the problem.
To find the time constant (τ) of the output voltage, we need to determine the equivalent resistance seen by the capacitor CTOT. Since the circuit only contains resistors, we can find the equivalent resistance using the formula:
1/RTOT = 1/R1 + 1/R2 + 1/R3
Substituting the given values for R1, R2, and R3:
1/RTOT = 1/1kΩ + 1/1kΩ + 1/2kΩ
Simplifying the equation:
1/RTOT = 2/1kΩ + 1/2kΩ
1/RTOT = 4/2kΩ + 2/2kΩ
1/RTOT = 6/2kΩ
1/RTOT = 3/1kΩ
RTOT = 1kΩ / 3
RTOT ≈ 333.33Ω
Now we can find the time constant (τ) using the formula:
τ = CTOT * RTOT
Substituting the values:
τ ≈ 83.3nF * 333.33Ω
τ ≈ 27.77μs
Therefore, the time constant (τ) of the output voltage is approximately 27.77μs.
Now let's find the output voltage (vO(t)) at time t=0.3ms.
Given that vO(0) = 1V, we can use the formula for the step response of a first-order system:
vO(t) = vO(0) * (1 - e^(-t/τ))
Substituting the values:
vO(t) = 1V * (1 - e^(-0.3ms/27.77μs))
vO(t) = 1V * (1 - e^(-0.3/27.77))
vO(t) ≈ 0.997V
Therefore, the output voltage (vO(t)) at time t=0.3ms is approximately 0.997V.
Finally, let's consider the step response where iI(t) = 1u(t)mA, vI(t) = 1u(t)V, and vO(0) = 0V.
Since the input is a step function, it will remain constant at 1u(t)mA and 1u(t)V for all values of time greater than or equal to 0.
For a step input, the output voltage will eventually settle to a steady-state value determined by the input. In this case, the steady-state output voltage can be found by substituting the value of the input into the transfer function of the circuit.
However, since the exact transfer function is not provided, I cannot provide a specific answer without further information.