Question

State and explain Kirchhoffs Laws in the current electricity.​

Answer 1

Kirchhoff's laws: (i) The algebraic sum of current meeting at an junction in a circuit is zero. In this law, the current flowing towards the junction are considered as +ve and those flowing away from the junction as −ve currents.

As shown in the figure, we have

$i1−i2−i3+i4−i5=0</p><p> \\ </p><p>or  i1+i4=i2+i3+i5 \\ \\ </p><p></p><p></p><p>$

(ii) In any closed mesh (or loop) of an electrical circuit, be algebraic sum of the product of the currents and resistances is equal to the total e.m.f. of the mesh.

If we go along the direction of conventional current, be potential difference will be taken as negative and opposite to it will be positive.

Inside the cell, if we move from low to high potential along the direction of conventional current the e.m.f. will be positive.

•For loop 1,

$\\ E2−i2R2−(i1+i2)R3=0</p><p></p><p> \\ or       E2=i2R2+(i1+i2)R3</p><p></p><p> \\ \\ </p><p>$

•For loop 2,

$\\ i </p><p>2</p><p> </p><p> R </p><p>2</p><p> </p><p> −E </p><p>2</p><p> </p><p> −i </p><p>1</p><p> </p><p> R </p><p>1</p><p> </p><p> +E </p><p>1</p><p> </p><p> =0</p><p> \\ or E </p><p>1</p><p> </p><p> −E </p><p>2</p><p> </p><p> =i </p><p>1</p><p> </p><p> R </p><p>1</p><p> </p><p> −i </p><p>2</p><p> </p><p> R </p><p>2</p><p> </p><p> </p><p></p><p> \\ </p><p>$

Answer 2

Answer:

Explanation:

The principle of this law is to conserve the electric charge. The law states that the amount of current flowing into a node is equal to the sum of currents flowing out of it. ... The total amount of energy gained is equal to the energy lost per unit charge.