- Home
- Standard 12
- Physics
Figure $(i)$ below shows a Wheatstone's bridge in which $P, Q, R$ and $S$ are fixed resistances, $G$ is a galvanometer and $B$ is a battery. For this particular case, the galvanometer shows zero deflection. Now, only the positions of $B$ and $G$ are interchanged, as shown in figure $(ii)$. The new deflection of the galvanometer
is to the left
is to the right
is zero
depends on the values of $P, Q, R$ and $S$
Solution
(c)
In case $(i)$, galvanometer shows zero deflection.
$\therefore V_B=V_D$
$\Rightarrow \frac{P}{S}=\frac{Q}{R} \quad \dots(i)$
When battery $B$ and galvanometer $G$ are interchanged, position of galvanometer is as shown below,
Now, ratio of resistances across galvanometer is
$\frac{S}{P} \text { and } \frac{R}{Q}$
As from Eq. $(i)$, $\frac{S}{P}=\frac{R}{Q}$
Hence, galvanometer still shows zero deflection because Wheatstone's bridge is balanced.
