Name: Kirchhoff's Second Law
Uploaded: 2019-07-19
Duration: 5 min 21 s
Description: Kirchhoff’s first law is very handy when calculating currents. Here, we explore his second potent law and use it to reduce complicated loops. Watch and learn.

Question

State Kirchhoff

'

s law for an electrical network. Using these laws deduce the condition for balance in wheatstone bridge.

Accepted Answer

In $1845$ , a German physicist, Kirchhoff developed a pair or set of rules or laws that addressed current and energy conservation within electrical circuits. These two rules are commonly known as Kirchhoff $'$ s current law $(KCL)$ while other law deals with the voltage sources present in a closed circuit, Kirchhoff $'$ s voltage law $(KVL)$ .
Kirchhoff $'$ s first law or $KCL$ states that the total current or charge entering a junction or node is exactly the same as the charge leaving the node because it has no other way to go but to exit, because no charge is lost within the node. At any intersection, the algebraic number of the currents is zero.

$∴ \sum_{n} I = 0$

Kirchhoff $'$ s second law, or $KVL$ , states that the total voltage around the loop is equal to the sum of all the voltage drops within the same loop in any closed loop network. Inside the loop the algebraic sum of all voltages must be equal to zero.

$∴ \sum_{n} IR + \underset{m}{\sum E = 0}$

Wheatstone bridge: Wheatstone $'$ s bridge circuit consists of four resistances $R_{1}, R_{2}, R_{3}, R_{4}$ are connected to form a closed path. A cell of emf $ε$ is connected between the point $A$ and $C$ and a galvanometer is connected between the points $D$ and $B$ . The current through the various branches are indicated in the figure. The current through the galvanometer is $I_{g}$ and the resistance if the galvanometer is $G$ .

There will be currents flowing across all the resistors as well as a current $I_{g}$ through $G$ . In the case of a balanced bridge, the resistors are such that $I_{g} = 0$ . In this case, Kirchhoff's junction rule is applied to junctions $D$ and $B$ .

Now, we apply Kirchhoff's loop rule to closed loops $ADBA$ and $CBDC$ .

The loop $ADBA$ gives,

$- I_{1} R_{1} + 0 + I_{2} R_{2} = 0 (∵ I_{g} = 0)$

$∴ I_{1} R_{1} = I_{2} R_{2}$

or $\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}} . . . (a)$

And the loop $CBDC$ gives,

$I_{4} R_{4} + 0 - I_{3} R_{3} = 0 (∵ I_{g} = 0)$

$∴ I_{3} R_{3} = I_{4} R_{4}$

or $\frac{I_{3}}{I_{4}} = \frac{R_{4}}{R_{3}} . . . (b)$

Now, let the current entering $A$ be $I$ then it divides into $I_{1}$ and $I_{2}$ such that, $I_{2} = I - I_{1}$ .

The current $I_{2}$ further divides into $I_{g}$ and $I_{4}$ such that $I_{4} = I_{2} - I_{g}$ .

In the case of a balanced bridge, $I_{g} = 0$ therefore, $I_{4} = I_{2}$ .

Similarly, we get $I_{3} = I_{1}$ .

Substituting $I_{3} = I_{1}$ and $I_{4} = I_{2}$ in the equation (b) we get,

$\frac{I_{1}}{I_{2}} = \frac{R_{4}}{R_{3}} . . . (c)$

From the equations $(a)$ and $(c)$ , we get,

$\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}} = \frac{R_{4}}{R_{3}}$

Hence, we obtain the condition,

$\frac{R_{2}}{R_{1}} = \frac{R_{4}}{R_{3}}$

which is the equation relating the four resistors called the balance condition for the galvanometer to give zero or null deflection.

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