Question
State Kirchhoff's laws of electric network.
✓
Answer
Kirchhoff’s First Law − Junction Rule
The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.
Let the currents be $I_1, I_2 I_3$, and $I_4$
Convention:
Current towards the junction - positive
Current away from the junction - negative
$I_3+\left(-I_1\right)+\left(-I_2\right)+\left(-I_4\right)=0$
Kirchhoff's Second Law - Loop Rule
In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them.
For closed part BACB, $E_1-E_2=I_1 R_1+I_2 R_2-I_3 R_3$
For closed part CADC, $E_2=I_3 R_3+I_4 R_4+I_5 R_5$
Wheatstone Bridge:
The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure. $R_1, R_2, R_3$ and $R_4$ are the four resistances.
Galvanometer (G) has a current $I_g$ flowing through it at balanced condition, $I_g=0$
Applying junction rule at B ,
$\therefore I_2=I_4$
Applying junction rule at D ,
$\therefore I_1=I_3$
Applying loop rule to closed loop ADBA,
$-I_1 R_1+0+I_2 R_2=0$
$\therefore \frac{I_1}{I_2}=\frac{R_2}{R_1} \ldots \ldots(1)$
Applying loop rule to closed loop CBDC,
$I_2 R_4+0-I_1 R_3=0 \quad \because I_3=I_1, I_4=I_2$
$\therefore \frac{I_1}{I_2}=\frac{R_4}{R_3} \cdots \cdots \text { (2) }$
|From equations (1) and (2),
$\frac{R_2}{R_1}=\frac{R_4}{R_3}$
This is the required balanced condition of Wheatstone Bridge.
The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.
Let the currents be $I_1, I_2 I_3$, and $I_4$
Convention:
Current towards the junction - positive
Current away from the junction - negative
$I_3+\left(-I_1\right)+\left(-I_2\right)+\left(-I_4\right)=0$
Kirchhoff's Second Law - Loop Rule
In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them.
For closed part BACB, $E_1-E_2=I_1 R_1+I_2 R_2-I_3 R_3$
For closed part CADC, $E_2=I_3 R_3+I_4 R_4+I_5 R_5$
Wheatstone Bridge:
The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure. $R_1, R_2, R_3$ and $R_4$ are the four resistances.
Galvanometer (G) has a current $I_g$ flowing through it at balanced condition, $I_g=0$
Applying junction rule at B ,
$\therefore I_2=I_4$
Applying junction rule at D ,
$\therefore I_1=I_3$
Applying loop rule to closed loop ADBA,
$-I_1 R_1+0+I_2 R_2=0$
$\therefore \frac{I_1}{I_2}=\frac{R_2}{R_1} \ldots \ldots(1)$
Applying loop rule to closed loop CBDC,
$I_2 R_4+0-I_1 R_3=0 \quad \because I_3=I_1, I_4=I_2$
$\therefore \frac{I_1}{I_2}=\frac{R_4}{R_3} \cdots \cdots \text { (2) }$
|From equations (1) and (2),
$\frac{R_2}{R_1}=\frac{R_4}{R_3}$
This is the required balanced condition of Wheatstone Bridge.
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