Question

State Kirchoff's law for an electrical network. Using these laws deduce the condition for balance in a wheatstone bridge.
Three resistors $2\Omega ,4\Omega$ and $5\Omega$ are combined in parallel. What is the total resistance of the combination?

Answer 1

Kirchoff's law,

1.Loop Law- The algebraic sum of voltage drop across each component is equal to the supplied voltage by the source.

2.Junction Law- The net (incoming + outgoing) current at each junction should be zero.

In case of Wheatstone bridge we know that the current through the galvanometer should be zero i.e $I_g=0$

Applying Loop law to closed loop ADBA

$-I_1{R}_1+0+I_2{R}_2=0$ ..... (1) because no source in this arm and the current in arm BD is zero.

For loop CBDC

$-I_2{R}_4+0+I_1{R}_3=0$ ...... (2)

From eqaution-1 $\frac{I_1}{I_2}=\frac{R_1}{R_2}$

From eqaution-1 $\frac{I_1}{I_2}=\frac{R_4}{R_3}$

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

Suppose the resulting resistance is $R$ then for parallel combination $\frac{1}{R}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}$ so $R=1.05\Omega$