CURRENT ELECTRICITY

Ans. Kirchhoff's First law:At any junction the sum of the currents entering the junction is equal to the sum of currents leaving the junction
$$\:\Sigma i\:=\:0$$

Kirchhoff's second law: The algebraic sum of potential difference around any closed loop is zero.
$$\Sigma V\:=\:0$$

Wheatstone bridge:-As shown in the figure $R_{1},R_{2},R_{3},R_{4}$, are resistors connected in the Wheatstone bridge circuit in the form of quadrilateral. Along one diagonal battery is connected, along the other diagonal galvanometer is connected.

According to Kirchhoff's First law at junctions A, B
$$\begin{array}\\{i=i_{1}+i_{2}}\\{i=i_{3}+i_{4}}\\\end{array}$$

And at junctions C, D respectively

$$\begin{array} {cc} i_{1}=i_g+i_3 \\ i_4=i_g+i_2 \end{array}$$

According to Kirchhoff's Second law in loops 'ADGCA' and 'BCGDB', potentia differences respectively
 $$\begin{array} -\:i_{1}R_{1}-i_{g}G\:+\:i_{2}R_{2}=\:0\\-\:i_{3}R_{3}+i_{g}G\:+\:i_{4}R_{4}=\:0\end{array}$$

When bridge is balanced $i_{g}=0$ from the above equations.

$$\frac{R_1}{R_2}=\frac{R_3}{R_4}$$

Therefore, the ratio of resistance of the adjacent side will always be equal.