To solve the problem, we need to apply Kirchhoff's loop law for the given circuit.
1. Kirchhoff's Loop Law:
Kirchhoff's loop law states that the sum of the potential differences (voltages) around any closed loop in a circuit must be zero. This is because energy is conserved in a closed loop.
2. Given Circuit Description:
In the given circuit:
- The current $I$ flows through a resistor of $3 \, \Omega$.
- The voltage of the battery is $12 \, \text{V}$.
3. Apply Kirchhoff’s Voltage Law:
The potential drop across the resistor is $3I$ (using Ohm's law: $V = IR$), and the voltage source provides $12 \, \text{V}$.
As per Kirchhoff's law, the sum of voltages in a closed loop is zero:
$ 3I - 12 = 0 $
Final Answer:
The equation for the given circuit as per Kirchhoff's loop law is $ {3I - 12 = 0} $.
