Question

Why is the PD of a battery less than its emf?

Answer 1

The potential difference (PD) of a battery is less than its electromotive force (emf) because of internal resistance. Here's a breakdown:

1. EMF is the total voltage a battery can provide when no current is flowing. It's the potential difference between the battery's terminals when it's open-circuited.
   

1. PD (Potential Difference) is the voltage across the battery's terminals when it's connected to a circuit and current is flowing. This is what you measure when the battery is in use.
   

Now, when current flows through the battery, some of the energy is used up overcoming the internal resistance of the battery. This internal resistance causes a voltage drop inside the battery itself, which reduces the voltage available at the terminals.

So, the PD is less than the emf because:

1. The battery has internal resistance (let's call it \( r \)).
   

1. When current \( I \) flows, the voltage drop across the internal resistance is \( I \times r \).
   

1. The voltage at the terminals (the PD) becomes:  
   

  \[
  \text{PD} = \text{EMF} - I \times r
  \]

In summary, the internal resistance of the battery causes a reduction in the voltage you get at the terminals when current flows, which is why the PD is lower than the emf.