(1) Find a formula for the capacitance based on the formula above. a circuit like the one in the above diagram with a resistance of 100 Ω is measured with an oscilloscope. The voltage amplitude on channel 1 was 1.52 V and on channel 2 was 192 mV. (2) Calculate the current amplitude based on the fact that channel 2 measures the voltage across just the 100 Ω resistor. The frequency measured was 19.98 kHz. (3) Calculate the angular frequency of the AC source. The angular frequency should appear in the formula you found in (1), along with other quantities like the voltage amplitude of the power source, which was 1.52 V measured on channel 1. (4) Based on the formula for capacitance that you found, the quantities you calculated, and the known resistance, calculate the capacitance in the circuit. Is it reasonably close to the 10 nF capacitance corresponding to the capacitor used? (5) When the frequency is small, what happens to the AC current in a capacitor? When the frequency is large, what happens to the AC current in a capacitor? What's the upper limit on current when a capacitor is in series with a resistor in AC and at what kinds of frequencies does it occur?
(1) Find a formula for the capacitance based on the formula above. a circuit like the one in the above diagram with a resistance of 100 Ω is measured with an oscilloscope. The voltage amplitude on channel 1 was 1.52 V and on channel 2 was 192 mV. (2) Calculate the current amplitude based on the fact that channel 2 measures the voltage across just the 100 Ω resistor. The frequency measured was 19.98 kHz. (3) Calculate the angular frequency of the AC source. The angular frequency should appear in the formula you found in (1), along with other quantities like the voltage amplitude of the power source, which was 1.52 V measured on channel 1. (4) Based on the formula for capacitance that you found, the quantities you calculated, and the known resistance, calculate the capacitance in the circuit. Is it reasonably close to the 10 nF capacitance corresponding to the capacitor used? (5) When the frequency is small, what happens to the AC current in a capacitor? When the frequency is large, what happens to the AC current in a capacitor? What's the upper limit on current when a capacitor is in series with a resistor in AC and at what kinds of frequencies does it occur?
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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(1) Find a formula for the capacitance based on the formula above. a circuit like the one in the above diagram with a resistance of 100 Ω is measured with an oscilloscope.
The voltage amplitude on channel 1 was 1.52 V and on channel 2 was 192 mV.
(2) Calculate the current amplitude based on the fact that channel 2 measures the voltage across just the 100 Ω resistor. The frequency measured was 19.98 kHz.
(3) Calculate the angular frequency of the AC source. The angular frequency should appear in the formula you found in (1), along with other quantities like the voltage amplitude of the power source, which was 1.52 V measured on channel 1.
(4) Based on the formula for capacitance that you found, the quantities you calculated, and the known resistance, calculate the capacitance in the circuit. Is it reasonably close to the 10 nF capacitance corresponding to the capacitor used?
(5) When the frequency is small, what happens to the AC current in a capacitor? When the frequency is large, what happens to the AC current in a capacitor? What's the upper limit on current when a capacitor is in series with a resistor in AC and at what kinds of frequencies does it occur?
![**Capacitors in AC**
The following diagram shows a circuit with an AC power source, a capacitor, and a resistor, with connections to the channels of an oscilloscope indicated.
**Diagram Explanation:**
- The diagram consists of an AC voltage source connected in series with a capacitor (C) and a resistor (R).
- There are two channels labeled:
- **Channel 1** is connected across the capacitor.
- **Channel 2** is connected across the resistor.
**Mathematical Explanation:**
Given an AC current \( I(t) = I \cos(\omega t) \) and source voltage \( V(t) = V \cos(\omega t + \phi) \), and that \( V(t) = V_C(t) + V_R(t) \), the following relationship is found for the circuit above:
\[
V \cos(\omega t + \phi) = \frac{I}{\omega C} \sin(\omega t) + IR \cos(\omega t)
\]
It can be shown that the following relationship holds for the amplitudes of each term in the above equation:
\[
V^2 = \left( \frac{I}{\omega C} \right)^2 + (IR)^2
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Transcribed Image Text:**Capacitors in AC**
The following diagram shows a circuit with an AC power source, a capacitor, and a resistor, with connections to the channels of an oscilloscope indicated.
**Diagram Explanation:**
- The diagram consists of an AC voltage source connected in series with a capacitor (C) and a resistor (R).
- There are two channels labeled:
- **Channel 1** is connected across the capacitor.
- **Channel 2** is connected across the resistor.
**Mathematical Explanation:**
Given an AC current \( I(t) = I \cos(\omega t) \) and source voltage \( V(t) = V \cos(\omega t + \phi) \), and that \( V(t) = V_C(t) + V_R(t) \), the following relationship is found for the circuit above:
\[
V \cos(\omega t + \phi) = \frac{I}{\omega C} \sin(\omega t) + IR \cos(\omega t)
\]
It can be shown that the following relationship holds for the amplitudes of each term in the above equation:
\[
V^2 = \left( \frac{I}{\omega C} \right)^2 + (IR)^2
\]
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