3. You are given voltage across capacitor in the above circuit for various time points. Which of the following plots would produce linear curve? In(V)us.time In(V)us.ln(time) In(Vo-V) vs. time In(Vo-V) vs. In(time)

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### Question 3: Analyzing Capacitor Voltage Plots to Identify Linear Curves

You are given voltage across a capacitor in the above circuit for various time points. Which of the following plots would produce a linear curve?

- [Selected] \( \ln(V_0) \) vs. time
- \( \ln(V_0) \) vs. \( \ln(\text{time}) \)
- \( \ln(V_0 - V) \) vs. time
- \( \ln(V_0 - V) \) vs. \( \ln(\text{time}) \)

### Explanation:
To determine which plot yields a linear curve, we need to understand the typical behavior of capacitor voltage over time in an electrical circuit. The options involve logarithmic transformations of voltage and time, which are often used to linearize exponential relationships. The correct choice indicates that plotting \( \ln(V_0) \) against time will produce a linear graph, suggesting an exponential decay of voltage over time.

### Visual Representation:
There are no graphs provided directly with the text, but here’s what each option typically represents:

1. **\( \ln(V_0) \) vs. time (Selected)**
   - A plot where the natural logarithm of the initial voltage \( V_0 \) is graphed against time. This is showing linearity assuming the voltage \( V_0 \) follows an exponential decay function without diminishing \( V \) over time.

2. **\( \ln(V_0) \) vs. \( \ln(\text{time}) \)**
   - This plot graphically represents how the initial voltage \( V_0 \) relates to the logarithm of time. This would be linear if \( V_0 \) followed a power-law decay rather than exponential.

3. **\( \ln(V_0 - V) \) vs. time**
   - Shows the natural logarithm of the voltage difference (initial minus present voltage) over time. This plot is often used to analyze RC circuits where \( \ln(V_0 - V) \) is linear with time due to the exponential nature of capacitor charging and discharging.

4. **\( \ln(V_0 - V) \) vs. \( \ln(\text{time}) \)**
   - Represents the relation between the logarithm of the voltage difference and the logarithm of time. This scenario typically suits systems that
Transcribed Image Text:### Question 3: Analyzing Capacitor Voltage Plots to Identify Linear Curves You are given voltage across a capacitor in the above circuit for various time points. Which of the following plots would produce a linear curve? - [Selected] \( \ln(V_0) \) vs. time - \( \ln(V_0) \) vs. \( \ln(\text{time}) \) - \( \ln(V_0 - V) \) vs. time - \( \ln(V_0 - V) \) vs. \( \ln(\text{time}) \) ### Explanation: To determine which plot yields a linear curve, we need to understand the typical behavior of capacitor voltage over time in an electrical circuit. The options involve logarithmic transformations of voltage and time, which are often used to linearize exponential relationships. The correct choice indicates that plotting \( \ln(V_0) \) against time will produce a linear graph, suggesting an exponential decay of voltage over time. ### Visual Representation: There are no graphs provided directly with the text, but here’s what each option typically represents: 1. **\( \ln(V_0) \) vs. time (Selected)** - A plot where the natural logarithm of the initial voltage \( V_0 \) is graphed against time. This is showing linearity assuming the voltage \( V_0 \) follows an exponential decay function without diminishing \( V \) over time. 2. **\( \ln(V_0) \) vs. \( \ln(\text{time}) \)** - This plot graphically represents how the initial voltage \( V_0 \) relates to the logarithm of time. This would be linear if \( V_0 \) followed a power-law decay rather than exponential. 3. **\( \ln(V_0 - V) \) vs. time** - Shows the natural logarithm of the voltage difference (initial minus present voltage) over time. This plot is often used to analyze RC circuits where \( \ln(V_0 - V) \) is linear with time due to the exponential nature of capacitor charging and discharging. 4. **\( \ln(V_0 - V) \) vs. \( \ln(\text{time}) \)** - Represents the relation between the logarithm of the voltage difference and the logarithm of time. This scenario typically suits systems that
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