It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit.

**Kirchhoffâ€™s Second rule**** ****(Voltage
rule or Loop rule)**

It states that in a
closed circuit the algebraic sum of the products of the current and resistance
of each part of the circuit is equal to the total emf included in the circuit.
This rule follows from the law of conservation of energy for an isolated system
(The energy supplied by the emf sources is equal to the sum of the energy
delivered to all resistors).The product of current and resistance is taken as
positive when the direction of the current is followed. Suppose if the
direction of current is opposite to the direction of the loop, then product of
current and voltage across the resistor is negative. It is shown in Figure 2.24
(a) and (b). The emf is considered positive when proceeding from the negative
to the positive terminal of the cell. It is shown in Figure 2.24 (c) and (d).

Kirchhoff voltage rule
has to be applied only when all currents in the circuit reach a steady state
condition (the current in various branches are constant).

**EXAMPLE 2.21**

The following figure
shows a complex network of conductors which can be divided into two closed
loops like ACE and ABC. Apply Kirchoffâ€™s voltage rule.

*Solution*

Thus applying Kirchoffâ€™s second law to the closed loop EACE

*I*_{1}*R*_{1}* *+* I*_{2}*R*_{2}* *+* I*_{3}*R*_{3}* *=* *Î¾

and for the closed loop ABCA

*I*_{4}*R*_{4}* *+* I*_{5}*R*_{5}-*I*_{2}*R*_{2}= 0

**EXAMPLE 2.22**

Calculate
the current that flows in the 1â€†Î© resistor in the following circuit.

*Solution*

We can
denote the current that flows from 9V battery as I_{1} and it splits
into I_{2} and I_{1 }â€“ I_{2}
in the junction according Kirchoffâ€™s current rule (KCR). It is shown below._{}

Now consider the loop EFCBE and apply KVR, we get

1I_{2}
+ 3I_{1} + 2I_{1} = 9

5I_{1} + I_{2} = 9 (1)

Applying KVR to the loop EADFE, we get

3 (I_{1}
â€“ I_{2} ) â€“ 1I_{2} = 6

3I_{1}
â€“ 4I_{2} = 6 (2)

Solving
equation (1) and (2), we get

I_{1}
= 1.83 A and I_{2} = -0.13 A

It
implies that the current in the 1 ohm resistor flows from F to E.

Tags : Explanation, Formulas, Solved Example Problems , 12th Physics : Current Electricity

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12th Physics : Current Electricity : Kirchhoffâ€™s Second rule (Voltage rule or Loop rule) | Explanation, Formulas, Solved Example Problems

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