As you all know, there are two Kirchhoff’s rules which are the junction rule and the loop rule. For instance, you see that a weightlifter needs to work to move a dumbbell from the ground (low potential) to his arm’s length (high potential). When he lets it go, gravity works to move it back to the ground. Thus, the dumbbell went in a loop. It began on the ground reached the maximum height then returned to the ground again. Thus, the net change in GPE (gravitational potential energy) is zero as there is neither gain nor loss of net energy. Similarly, we have a battery doing the same thing. However, in this case, it moves from negative to positive. Thus, this change in electric potential will be called voltage. We will study here about the kirchhoff’s loop rule formula.

**Definition**

Kirchhoff’s loop rule explains that the sum of all the electric potential differences nearby a loop is 0. Sometimes, we also refer to it as Kirchhoff’s voltage law or Kirchhoff’s second law. In other words, it states that the energy which the battery supplies get used up by all the other components in a loop.

It is so because energy cannot enter or leave a closed circuit. The rule is an application of the preservation of energy in terms of the electric potential difference \(\Delta V\).

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**Formula**

You will notice that any loop of a closed circuit consists of a number of circuit elements like batteries and resistors. The sum of the voltage differences throughout all these elements of circuits should be zero. We refer to this as Kirchhoff’s Loop Rule.

We measure the differences in voltage in Volts (V). When you have the current I in the loop given Amperes (A) and resistance of circuit elements in Ohms (Ω), then we can find the voltage difference across a resistor by using the formula V = IR.

Thus, you will get:

\(\sum V\) = 0

Over here:

\(\sum V\) is the sum of voltage differences around a circuit loop which is 0

V is the voltage difference (Volts-V)

## Solved Examples

**Question- **We have a circuit loop which comprises of 3 resistors and a voltage source which is the battery. The current in the loop is I = +4.00 A, in a clockwise direction. The voltage supplied by the battery is v_{b} = 100.0 V. Furthermore, the resistance values of the two resistors out of the three are R_{1 }is 10. 0 Ω and R_{2 }is 8.00 Ω. You need to find the value of the resistor R_{3}.

**Solution- **

The Kirchhoff’s Loop Rule says that the sum of the voltage differences near the loop should be 0. Therefore, in order to find the sum, we need to choose the direction of travel. The direction of the positive current is said to be clockwise and thus, we will use this as the direction of travel so we can find out the sum. The voltage source i.e. the battery in the problem is said to have a positive voltage value in clockwise direction. The voltage drops in this direction due to the three resistors. We see that the magnitudes of the voltage drops are identical to the resistance multiplied by the current in the loop. Therefore, the sum of the voltage differences will be:

\(\sum V\) = 0

\(\sum V\) = \(V_{B} – 1R_{1} – 1R_{2} – 1R_{3}\) = 0

∴ \(V_{B} – 1R_{1} – 1R_{2} – 1R_{3}\) = 0

(100.0 V) – (4.00 A) (10.0 Ω) – (4.00 A) (8.00 Ω) – (4.00 A) \(R_{3}\) = 0

100.0 V – 40.0 V – 32.0 V – (4.00 A) \(R_{3}\) = 0

28.0 V – (4.00 A) \(R_{3}\) = 0

We can find the value of the third resistor by rearranging the formula:

28.0 V – (4.00 A) \(R_{3}\) = 0

∴ – (4.00 A) \(R_{3}\) = -28.0 V

∴ (4.00 A) \( R_{3}\) = 28.0 V

∴ \(R_{3}\) = \(\frac{28.0V}{4.00A}\)

∴ \(R_{3}\) = 7.00 \(\frac{V}{A}\)

∴ \(R_{3}\) = 7.00 Ω

Therefore, the value of resistor \(R_{3}\) comes out as 7.00 Ω (Ohms).

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