Question

Obtain the balancing  condition for the Wheatstone bridge arrangements as shown in Figure 4 below:

Answer 1

Let ${\displaystyle I_3}$ and ${\displaystyle I_4}$  be the currents in resistors Q and S respectively . Let ${\displaystyle I_g}$ be the current through galvanometer. For balanced condition, 

${\displaystyle I_g=0}$

Applying junction law at ‘b’ we get

${\displaystyle I_1=I_3+I_g}$

${\displaystyle \because I_g=0,I_1=I_3}$    ....(i)

Applying junction law at ‘d’, we get

${\displaystyle I_2+I_g=I_4}$

${\displaystyle \because I_g=0,I_2=I_4}$    ....(ii)

Applying loop law in the loop abda, we get

${\displaystyle -I_1·P-I_g·Q+-I_2·R=0}$

⇒ ${\displaystyle -I_{1}P+I_{2}R=0}$  (${\displaystyle \because I_g=0}$)

⇒ ${\displaystyle I_{1}P=I_{2}R}$

⇒ ${\displaystyle \frac{P}{R}=\frac{I_2}{I_1}}$               ....(iii)

Applying loop law in the loop bcdb, we get

${\displaystyle -I_3·Q+I_4·S+I_g·6=0}$

⇒ ${\displaystyle -I_3·Q+I_4·S+0=0 (\because I_g=0)}$

⇒ ${\displaystyle -I_{3}Q=I_{4}S}$

⇒ ${\displaystyle \frac{Q}{S}=\frac{I_4}{I_3}}$

⇒ ${\displaystyle \frac{Q}{S}=\frac{I_2}{I_1}}$             ...(iv) [using eq.(i) and (ii)]

From eq. (iii) and (iv), ${\displaystyle \frac{P}{R}=\frac{Q}{s}}$

⇒ ${\displaystyle \frac{P}{Q}=\frac{R}{S}}$

This is the balanced condition.