**Question:**If the switch has been closed for a time period long enough for the capacitor to become fully charged, and then the switch is opened, how long before the current through resistor \( R_1 \) reaches half of its initial value?**Hint for (d)**In time \( t = \) (input area for time in ms), the current through resistor \( R_1 \) reaches half its initial value. ### Electrical Circuit AnalysisConsider the following electrical circuit diagram:**Circuit Components:**- **Voltage Source (V1):** 24 V- **Resistor 1 (R1):** 10 kΩ- **Resistor 2 (R2):** 30 kΩ- **Capacitor (C1):** 10 µF**Circuit Description:**1. The voltage source \( V_1 \) of 24 V is connected in series with a switch.2. When the switch is closed, the circuit branches into two parallel paths: - The first branch contains Resistor \( R_1 \) with a resistance of 10 kΩ. - The second branch contains both Resistor \( R_2 \) with a resistance of 30 kΩ and Capacitor \( C_1 \) with a capacitance of 10 µF, connected in series. **Assumptions:**- The capacitor \( C_1 \) is initially uncharged.This setup is often analyzed to understand the transient response of the capacitor and the steady-state behavior of the circuit. When the switch is closed, it is important to solve for the voltage across the capacitor and the current flowing through the resistors as functions of time.### Understanding Transients in the RC CircuitWhen the switch is initially closed, the capacitor begins to charge. The voltage across the capacitor as a function of time ( \( V_C(t) \) ) can be derived using Kirchhoff's voltage law (KVL) and the formulas for charging a capacitor in RC circuits.\[ V_C(t) = V_s \left(1 - e^{-\frac{t}{R_{\text{eq}}C}} \right) \]where:- \( V_s \) is the supply voltage (24 V in this case).- \( R_{\text{eq}} \) is the equivalent resistance seen by the capacitor.- \( t \) is time.- \( C \) is the capacitance (10 µF in this case).#### Equivalent Resistance:Given that \( R_1 \) and \( R_2 \) are in series:\[ R_{\text{eq}} = R1 + R2 \]Substitute the values:\[ R_{\text{eq}} = 10 kΩ + 30 kΩ = 40 kΩ \]Now the time constant ( \( \tau \

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d. If the switch has been closed for a time period long enough for the capacitor to become fully charged, and then the switch is opened, how long before the current through resistor R₁ reaches half of its initial value? Hint for (d) In time t = initial value. ms, the current through resistor R₁ reaches half its

d. If the switch has been closed for a time period long enough for the capacitor to become fully charged, and then the switch is opened, how long before the current through resistor R₁ reaches half of its initial value? Hint for (d) In time t = initial value. ms, the current through resistor R₁ reaches half its

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ISBN:9781305952300

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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### Electrical Circuit Analysis

Consider the following electrical circuit diagram:

**Circuit Components:**
- **Voltage Source (V1):** 24 V
- **Resistor 1 (R1):** 10 kΩ
- **Resistor 2 (R2):** 30 kΩ
- **Capacitor (C1):** 10 µF

**Circuit Description:**
1. The voltage source \( V_1 \) of 24 V is connected in series with a switch.
2. When the switch is closed, the circuit branches into two parallel paths:
- The first branch contains Resistor \( R_1 \) with a resistance of 10 kΩ.
- The second branch contains both Resistor \( R_2 \) with a resistance of 30 kΩ and Capacitor \( C_1 \) with a capacitance of 10 µF, connected in series.

**Assumptions:**
- The capacitor \( C_1 \) is initially uncharged.

This setup is often analyzed to understand the transient response of the capacitor and the steady-state behavior of the circuit. When the switch is closed, it is important to solve for the voltage across the capacitor and the current flowing through the resistors as functions of time.

### Understanding Transients in the RC Circuit

When the switch is initially closed, the capacitor begins to charge. The voltage across the capacitor as a function of time ( \( V_C(t) \) ) can be derived using Kirchhoff's voltage law (KVL) and the formulas for charging a capacitor in RC circuits.

\[ V_C(t) = V_s \left(1 - e^{-\frac{t}{R_{\text{eq}}C}} \right) \]

where:

- \( V_s \) is the supply voltage (24 V in this case).
- \( R_{\text{eq}} \) is the equivalent resistance seen by the capacitor.
- \( t \) is time.
- \( C \) is the capacitance (10 µF in this case).

#### Equivalent Resistance:

Given that \( R_1 \) and \( R_2 \) are in series:

\[ R_{\text{eq}} = R1 + R2 \]

Substitute the values:

\[ R_{\text{eq}} = 10 kΩ + 30 kΩ = 40 kΩ \]

Now the time constant ( \( \tau \

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