
Given:
A battery of emf \( E \) and internal resistance \( r \) is connected in series with an external resistance \( R \).
Concept:
The voltage across the external resistance \( R \) is given by: \[ V = E - Ir \] where \( I = \frac{E}{R + r} \) So the voltage across \( R \) becomes: \[ V = E - \left( \frac{E}{R + r} \right) r = \frac{E R}{R + r} \]
For the battery to act as a constant voltage source:
The voltage across the external resistance \( V \approx E \). This happens when the drop across internal resistance is negligible. \[ \text{This implies } r \ll R \]
Therefore, the battery will act as a constant voltage source when:
r << R
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:

A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of:
