7.6 Step Response pada Rangkaian RC (3 dari 3)

Summary

This video tutorial guides viewers through the analysis of a step response in an RC circuit. The scenario involves a switch moving between two positions, affecting the voltage across a capacitor. The instructor first calculates the voltage at different time intervals, starting with the voltage before the switch is flipped, followed by the voltage immediately after the switch is moved, and finally the voltage at a later time. The step response is modeled using the capacitor's voltage equation, with a focus on understanding the behavior of the circuit over time. The video offers a clear, step-by-step explanation of how to approach and solve this type of circuit problem.

Takeaways

• 😀 The circuit starts with a blue switch in position 1, where the capacitor is fully charged and no current flows through the circuit.
• 😀 At time t = 0, the switch moves to position 2, and the goal is to analyze the step response of the circuit for t > 0.
• 😀 Prior to t = 0, the voltage across the capacitor is the same as the voltage across the resistor (4kΩ) due to the voltage divider principle.
• 😀 The voltage across the capacitor just before t = 0 is calculated to be 8V, using a voltage divider between the 6kΩ resistor and the 4kΩ resistor.
• 😀 The capacitor voltage does not change abruptly when the switch moves to position 2, as capacitors maintain their voltage instantly.
• 😀 The voltage at t = 0 is 8V, the same as it was just before t = 0, as the capacitor's voltage remains unchanged.
• 😀 After t = 0, the circuit behaves with the 2kΩ resistor in place, and the time constant (τ) is calculated to be 2 seconds using τ = RC.
• 😀 For t > 0, the capacitor gradually discharges or charges toward the source voltage (10V), with the voltage following an exponential curve.
• 😀 The steady-state voltage across the capacitor after a long time (t → ∞) is equal to the source voltage, 10V, because the capacitor acts as an open circuit.
• 😀 The step response of the voltage across the capacitor is given by the formula: V(t) = 10V - 2V * e^(-0.5t). At t = 2 seconds, the voltage is calculated to be approximately 8.6V.

Q & A

What is the main objective of the script in terms of the RC circuit analysis?

The main objective is to determine the step response (VT) of an RC circuit after the switch changes positions, as well as to calculate the voltage across the capacitor at t = 2 seconds.

What happens to the circuit when the switch moves from position 1 to position 2?

When the switch moves from position 1 to position 2, the voltage across the capacitor does not change suddenly. The capacitor retains its initial voltage just before the switch is moved.

How is the initial voltage across the capacitor (V0) calculated before t = 0?

Before t = 0, the initial voltage across the capacitor is calculated using the voltage divider rule. The voltage across the resistor is 8V, which is the initial voltage across the capacitor.

What is the significance of the time constant (τ) in an RC circuit?

The time constant (τ) represents the time it takes for the capacitor to charge or discharge to approximately 63% of its maximum voltage. It is calculated as the product of the resistance (R) and capacitance (C).

What is the value of the time constant (τ) in this particular circuit?

In this circuit, the time constant (τ) is calculated as τ = R * C = 2 kΩ * 1 µF = 2 seconds.

What happens to the capacitor when a long time has passed after the switch moves to position 2?

After a long time, the capacitor is fully charged and behaves like an open switch, meaning no current flows through it. The voltage across the capacitor will stabilize at the source voltage (10V).

What is the general form of the equation for the step response in this RC circuit?

The general form of the equation for the step response is VT(t) = V_s * (1 - e^(-t/τ)), where V_s is the source voltage, and τ is the time constant.

What is the specific equation for VT(t) in this case, considering the initial conditions?

In this case, the equation for VT(t) is VT(t) = 10V * (1 - e^(-0.5t)), where 10V is the source voltage, and the time constant is 2 seconds.

How do you calculate the voltage across the capacitor at t = 2 seconds?

At t = 2 seconds, the voltage across the capacitor is calculated by substituting t = 2 into the equation VT(t) = 10 - 2 * e^(-0.5 * 2). The result is approximately 8.604V.

What does the graph of the voltage across the capacitor look like over time?

The graph shows an exponential increase in voltage from 0V to 10V. At t = 2 seconds, the voltage is approximately 8.604V, and the voltage gradually approaches the steady-state value of 10V.