1 Answer
To prove **Ohm's Law**, which states:
$$
V = IR
$$
where:
* $V$ is the **voltage** (potential difference),
* $I$ is the **current**, and
* $R$ is the **resistance**,
you can approach it **experimentally** or **theoretically**.
---
### Experimental Proof:
Ohm's Law is based on **experimental observation**. Here's how it's demonstrated:
1. **Set up a simple circuit** with:
* A variable power supply (voltage source),
* A known resistor $R$,
* An ammeter (to measure current $I$),
* A voltmeter (to measure voltage $V$) across the resistor.
2. **Vary the voltage** gradually and record the corresponding current through the resistor.
3. **Plot $V$ vs. $I$**:
* The graph should be a straight line.
* The slope of the line is the resistance $R$.
$$
\text{Slope} = \frac{V}{I} = R \Rightarrow V = IR
$$
---
### Theoretical Justification:
From the microscopic view of current flow in a conductor:
* Electrons move under an electric field $E$,
* The drift velocity $v_d$ is proportional to $E$,
* Current density $J = nqv_d$, where $n$ is charge carrier density, $q$ is charge,
* Ohm’s Law in differential form: $\vec{J} = \sigma \vec{E}$,
* Integrating over a conductor of length $L$ and cross-sectional area $A$:
$$
V = EL, \quad I = JA = \sigma EA
$$
So:
$$
V = \frac{I}{\sigma A}L = IR \quad \text{(where } R = \frac{L}{\sigma A} \text{)}
$$
This shows the proportionality of voltage to current for a given resistance.
---
Would you like a diagram of the experimental setup or graph of the $V$-$I$ relationship?
$$
V = IR
$$
where:
* $V$ is the **voltage** (potential difference),
* $I$ is the **current**, and
* $R$ is the **resistance**,
you can approach it **experimentally** or **theoretically**.
---
### Experimental Proof:
Ohm's Law is based on **experimental observation**. Here's how it's demonstrated:
1. **Set up a simple circuit** with:
* A variable power supply (voltage source),
* A known resistor $R$,
* An ammeter (to measure current $I$),
* A voltmeter (to measure voltage $V$) across the resistor.
2. **Vary the voltage** gradually and record the corresponding current through the resistor.
3. **Plot $V$ vs. $I$**:
* The graph should be a straight line.
* The slope of the line is the resistance $R$.
$$
\text{Slope} = \frac{V}{I} = R \Rightarrow V = IR
$$
---
### Theoretical Justification:
From the microscopic view of current flow in a conductor:
* Electrons move under an electric field $E$,
* The drift velocity $v_d$ is proportional to $E$,
* Current density $J = nqv_d$, where $n$ is charge carrier density, $q$ is charge,
* Ohm’s Law in differential form: $\vec{J} = \sigma \vec{E}$,
* Integrating over a conductor of length $L$ and cross-sectional area $A$:
$$
V = EL, \quad I = JA = \sigma EA
$$
So:
$$
V = \frac{I}{\sigma A}L = IR \quad \text{(where } R = \frac{L}{\sigma A} \text{)}
$$
This shows the proportionality of voltage to current for a given resistance.
---
Would you like a diagram of the experimental setup or graph of the $V$-$I$ relationship?