What is the formula for galvanometer to voltmeter?
2 Answers
To convert a **galvanometer** into a **voltmeter**, a **series resistor** (also called a **multiplier resistor** or **high resistance**) is added to the galvanometer. This allows the galvanometer to measure higher voltages across a circuit.
### Formula for the conversion:
The formula to determine the value of the series resistor \( R_s \) to convert the galvanometer into a voltmeter is given by:
\[
R_s = \frac{V}{I_g} - R_g
\]
Where:
- \( R_s \) is the series resistor (multiplier) to be added.
- \( V \) is the full-scale voltage (the maximum voltage that the voltmeter should measure).
- \( I_g \) is the full-scale deflection current of the galvanometer (i.e., the current needed for the galvanometer to show full deflection).
- \( R_g \) is the internal resistance of the galvanometer.
### Explanation of Terms:
- **Galvanometer**: A device that measures small currents. It typically has a very low resistance and is very sensitive to small currents.
- **Voltmeter**: A device used to measure voltage across a circuit.
- **Series Resistor**: The resistor added in series with the galvanometer to limit the current, allowing the galvanometer to measure higher voltages.
### How it Works:
1. The galvanometer alone can only measure small voltages because it is designed to detect small currents. If you apply a high voltage across the galvanometer directly, it may damage the device because it’s not designed for high currents or voltages.
2. By adding a series resistor, you increase the total resistance of the circuit, reducing the current flowing through the galvanometer.
3. This allows the galvanometer to measure a higher voltage without exceeding its current limit.
### Example:
Let's say you have:
- A galvanometer with a full-scale deflection current \( I_g = 1 \, \text{mA} \) (0.001 A),
- A galvanometer resistance \( R_g = 50 \, \Omega \),
- You want the voltmeter to measure up to \( V = 10 \, \text{V} \).
Using the formula:
\[
R_s = \frac{V}{I_g} - R_g = \frac{10}{0.001} - 50 = 10,000 - 50 = 9,950 \, \Omega
\]
So, you would need to add a series resistor of \( 9,950 \, \Omega \) to convert the galvanometer into a voltmeter that can measure up to 10V.
This method ensures that the galvanometer's deflection corresponds to the correct voltage reading.
### Formula for the conversion:
The formula to determine the value of the series resistor \( R_s \) to convert the galvanometer into a voltmeter is given by:
\[
R_s = \frac{V}{I_g} - R_g
\]
Where:
- \( R_s \) is the series resistor (multiplier) to be added.
- \( V \) is the full-scale voltage (the maximum voltage that the voltmeter should measure).
- \( I_g \) is the full-scale deflection current of the galvanometer (i.e., the current needed for the galvanometer to show full deflection).
- \( R_g \) is the internal resistance of the galvanometer.
### Explanation of Terms:
- **Galvanometer**: A device that measures small currents. It typically has a very low resistance and is very sensitive to small currents.
- **Voltmeter**: A device used to measure voltage across a circuit.
- **Series Resistor**: The resistor added in series with the galvanometer to limit the current, allowing the galvanometer to measure higher voltages.
### How it Works:
1. The galvanometer alone can only measure small voltages because it is designed to detect small currents. If you apply a high voltage across the galvanometer directly, it may damage the device because it’s not designed for high currents or voltages.
2. By adding a series resistor, you increase the total resistance of the circuit, reducing the current flowing through the galvanometer.
3. This allows the galvanometer to measure a higher voltage without exceeding its current limit.
### Example:
Let's say you have:
- A galvanometer with a full-scale deflection current \( I_g = 1 \, \text{mA} \) (0.001 A),
- A galvanometer resistance \( R_g = 50 \, \Omega \),
- You want the voltmeter to measure up to \( V = 10 \, \text{V} \).
Using the formula:
\[
R_s = \frac{V}{I_g} - R_g = \frac{10}{0.001} - 50 = 10,000 - 50 = 9,950 \, \Omega
\]
So, you would need to add a series resistor of \( 9,950 \, \Omega \) to convert the galvanometer into a voltmeter that can measure up to 10V.
This method ensures that the galvanometer's deflection corresponds to the correct voltage reading.
To convert a galvanometer into a voltmeter, you need to add a resistor in series with the galvanometer. The process is based on the concept of using the galvanometer's deflection to measure voltage.
### Components Involved
1. **Galvanometer (G)**: An instrument that measures small currents. It has a specific full-scale deflection (FSD) for a given current \( I_g \).
2. **Shunt Resistor (R_s)**: A resistor that is added in series to the galvanometer to limit the current passing through it.
3. **Series Resistor (R)**: A resistor added to convert the galvanometer to a voltmeter.
### Basic Principles
- The galvanometer is calibrated to measure current \( I_g \), which corresponds to a certain deflection on the scale.
- When you connect a voltage \( V \) across the series combination of the galvanometer and the resistor \( R \), the current flowing through the circuit will be the total current \( I \).
### Key Formulas
1. **Ohm's Law**:
\[
V = I \times R_{\text{total}}
\]
where \( R_{\text{total}} = R_g + R \) (the internal resistance of the galvanometer plus the added resistor).
2. **Current Division**: The current through the galvanometer can be expressed as:
\[
I_g = \frac{V}{R + R_g}
\]
3. **Voltage across the galvanometer**:
\[
V_g = I_g \times R_g
\]
### Designing the Voltmeter
To convert the galvanometer into a voltmeter:
1. Determine the maximum voltage \( V \) you want to measure.
2. Use the galvanometer’s specifications to find the full-scale deflection current \( I_g \) and its resistance \( R_g \).
3. Rearranging the formula for \( V_g \):
\[
V = I_g \times (R + R_g)
\]
Hence, if you know \( I_g \) and \( R_g \), you can calculate the required series resistor \( R \):
\[
R = \frac{V}{I_g} - R_g
\]
### Example
1. Assume you have a galvanometer with:
- Full-scale deflection current \( I_g = 1 \, \text{mA} \)
- Internal resistance \( R_g = 100 \, \Omega \)
2. You want to convert it to measure a maximum voltage of \( V = 10 \, \text{V} \):
\[
R = \frac{10 \, \text{V}}{0.001 \, \text{A}} - 100 \, \Omega = 10000 \, \Omega - 100 \, \Omega = 9900 \, \Omega
\]
3. Connect a \( 9900 \, \Omega \) resistor in series with the galvanometer. The resulting circuit will allow you to measure voltages up to \( 10 \, \text{V} \).
### Conclusion
By calculating the proper series resistance, you can effectively utilize a galvanometer as a voltmeter. This method ensures that the galvanometer operates within its safe current range while allowing for a wider voltage measurement capability.
### Components Involved
1. **Galvanometer (G)**: An instrument that measures small currents. It has a specific full-scale deflection (FSD) for a given current \( I_g \).
2. **Shunt Resistor (R_s)**: A resistor that is added in series to the galvanometer to limit the current passing through it.
3. **Series Resistor (R)**: A resistor added to convert the galvanometer to a voltmeter.
### Basic Principles
- The galvanometer is calibrated to measure current \( I_g \), which corresponds to a certain deflection on the scale.
- When you connect a voltage \( V \) across the series combination of the galvanometer and the resistor \( R \), the current flowing through the circuit will be the total current \( I \).
### Key Formulas
1. **Ohm's Law**:
\[
V = I \times R_{\text{total}}
\]
where \( R_{\text{total}} = R_g + R \) (the internal resistance of the galvanometer plus the added resistor).
2. **Current Division**: The current through the galvanometer can be expressed as:
\[
I_g = \frac{V}{R + R_g}
\]
3. **Voltage across the galvanometer**:
\[
V_g = I_g \times R_g
\]
### Designing the Voltmeter
To convert the galvanometer into a voltmeter:
1. Determine the maximum voltage \( V \) you want to measure.
2. Use the galvanometer’s specifications to find the full-scale deflection current \( I_g \) and its resistance \( R_g \).
3. Rearranging the formula for \( V_g \):
\[
V = I_g \times (R + R_g)
\]
Hence, if you know \( I_g \) and \( R_g \), you can calculate the required series resistor \( R \):
\[
R = \frac{V}{I_g} - R_g
\]
### Example
1. Assume you have a galvanometer with:
- Full-scale deflection current \( I_g = 1 \, \text{mA} \)
- Internal resistance \( R_g = 100 \, \Omega \)
2. You want to convert it to measure a maximum voltage of \( V = 10 \, \text{V} \):
\[
R = \frac{10 \, \text{V}}{0.001 \, \text{A}} - 100 \, \Omega = 10000 \, \Omega - 100 \, \Omega = 9900 \, \Omega
\]
3. Connect a \( 9900 \, \Omega \) resistor in series with the galvanometer. The resulting circuit will allow you to measure voltages up to \( 10 \, \text{V} \).
### Conclusion
By calculating the proper series resistance, you can effectively utilize a galvanometer as a voltmeter. This method ensures that the galvanometer operates within its safe current range while allowing for a wider voltage measurement capability.