**Kirchhoff's First Rule:**

At any junction of several circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it.

In the above junction, current I enters it and currents I

_{1}and I

_{2}leave it. Then,

I = I

_{1}+ I2

This is a consequence of charge conservation and assumption that currents are steady, that is no charge piles up at the junction.

**Kirchhoff's Second Rule:**

The algebraic sum of changes in potential around any closed resistor loop must be zero. This is based on the principle that electrostatic forces alone cannot do any work in a closed loop, since this work is equal to potential difference, which is zero, if we start at one point of the loop and come back to it.

This gives: (R

_{1}+ R

_{2}) I

_{1}+ R

_{3}I

_{1}+ R

_{4}I

_{4}= 0

**In case of current loops:**

i) Choose any closed loop in the network and designate a direction (in this example counter clockwise) to traverse the loop.

ii) Go around the loop in the designated direction, adding emf's and potential differences. An emf is counted as positive when it is traversed (-) to (+) and negative in the opposite case i.e., from (+) to (-). An IR term is counted negative if the resistor is traversed in the same direction of the assumed current, and positive if in the opposite direction.

iii) Equate the total sum to zero.

**Wheatstone Bridge:**

Wheatstone bridge is an arrangement of four resistances R

_{1}, R

_{2}, R

_{3}, R

_{4}. The null point condition is given by,

This is also known as the balanced condition. If

R

_{1}, R

_{2}, R

_{3}are known, R

_{4}can be determined

In a balanced condition of the meter bridge,

*s*is the resistance per unit length of wire and l

_{1}is the length of wire from one end where null point is obtained.

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