Kirchoff's second law

The algebraic sum of the EMFs is equal to the sum of potential differences around a closed loop.

This essentially means that:

• the total EMF (voltage) supplied by the power source equals the total potential difference (voltage) across the components in the loop.

We can write this as:

\sum \epsilon = \sum V

Example

In a simple circuit, a battery of EMF 12V is connected in series with two resistors of resistance 2\Omega and 4\Omega. Calculate the potential difference across each resistor.

• Total resistance = 2\Omega + 4\Omega = 6\Omega
• Total EMF = 12V
  • so the total potential difference across the resistors must also be 12V.
• V=IR
  • V=12V
  • R=6\Omega
  • I=\frac{V}{R}=\frac{12V}{6\Omega}=2A
• Potential difference across 2\Omega resistor:
  • V=IR=2A \times 2\Omega=4V
• Potential difference across 4\Omega resistor:
  • V=IR=2A \times 4\Omega=8V