Voltage is a crucial parameter in electrical systems, reflecting the potential difference that drives current through a circuit. This article explains how to find the voltage (\( V \)) when capacitance (\( C \)) and charge (\( Q \)) are known, using the formula \( V = \dfrac{Q}{C} \). We’ll provide three practical examples to illustrate the calculations.
Formula to Determine Voltage
Voltage (\( V \)) can be calculated using the formula:
\[ V = \dfrac{Q}{C} \]
where:
- \( V \) is the voltage (in volts, V),
- \( Q \) is the charge (in coulombs, C),
- \( C \) is the capacitance (in farads, F).
Example 1: Voltage in a Camera Flash Capacitor
Scenario: A camera flash capacitor has a charge of \( 0.05 \, \text{C} \) and a capacitance of \( 250 \, \mu\text{F} \). What is the voltage?
Step-by-Step Calculation:
1. Given:
\[ Q = 0.05 \, \text{C} \]
\[ C = 250 \, \mu\text{F} = 250 \times 10^{-6} \, \text{F} \]
2. Substitute Values into the Voltage Formula:
\[ V = \dfrac{Q}{C} \]
\[ V = \dfrac{0.05}{250 \times 10^{-6}} \]
3. Perform the Calculation:
\[ V = 200 \, \text{V} \]
Final Value
The voltage in the camera flash capacitor is:
\[ V = 200 \, \text{V} \]
Example 2: Voltage in a Car Battery Capacitor
Scenario: A car battery capacitor has a charge of \( 0.03 \, \text{C} \) and a capacitance of \( 4700 \, \mu\text{F} \). Calculate the voltage.
Step-by-Step Calculation:
1. Given:
\[ Q = 0.03 \, \text{C} \]
\[ C = 4700 \, \mu\text{F} = 4700 \times 10^{-6} \, \text{F} \]
2. Substitute Values into the Voltage Formula:
\[ V = \dfrac{Q}{C} \]
\[ V = \dfrac{0.03}{4700 \times 10^{-6}} \]
3. Perform the Calculation:
\[ V \approx 6.38 \, \text{V} \]
Final Value
The voltage in the car battery capacitor is approximately:
\[ V \approx 6.38 \, \text{V} \]
Example 3: Voltage in a Power Supply Capacitor
Scenario: A power supply capacitor has a charge of \( 0.015 \, \text{C} \) and a capacitance of \( 330 \, \mu\text{F} \). What is the voltage?
Step-by-Step Calculation:
1. Given:
\[ Q = 0.015 \, \text{C} \]
\[ C = 330 \, \mu\text{F} = 330 \times 10^{-6} \, \text{F} \]
2. Substitute Values into the Voltage Formula:
\[ V = \dfrac{Q}{C} \]
\[ V = \dfrac{0.015}{330 \times 10^{-6}} \]
3. Perform the Calculation:
\[ V \approx 45.45 \, \text{V} \]
Final Value
The voltage in the power supply capacitor is approximately:
\[ V \approx 45.45 \, \text{V} \]
Summary
To find the voltage (\( V \)) given the charge (\( Q \)) and capacitance (\( C \)), use the formula:
\[ V = \dfrac{Q}{C} \]
In the examples provided:
1. A camera flash capacitor with \( 0.05 \, \text{C} \) and \( 250 \, \mu\text{F} \) has a voltage of \( 200 \, \text{V} \).
2. A car battery capacitor with \( 0.03 \, \text{C} \) and \( 4700 \, \mu\text{F} \) has a voltage of approximately \( 6.38 \, \text{V} \).
3. A power supply capacitor with \( 0.015 \, \text{C} \) and \( 330 \, \mu\text{F} \) has a voltage of approximately \( 45.45 \, \text{V} \).
