Physics, asked by sahilbhor628, 9 months ago

A Wheatstone’s bridge ABCD is balanced with a galvanometer between
      the points B and D. At balance, thepotential at B and D

Answers

Answered by adityabhatt205
5

Answer:

V = 0

Explanation:

For a balanced Wheatstone bridge ABCD, the current passing through B & D will be 0 A.

Current will be 0 A only if Voltage (Potential Difference) = 0

Hope that helps...

Answered by ravilaccs
0

Answer:

The potential between B and D is +33 V

Explanation:

Given: Wheatstone’s bridge ABCD is balanced with a galvanometer between the points B and D

To find: Potential between B and D

Solution:

(a) Current through an arm $A B C$,

$$\begin{aligned}&I=\frac{48}{10+20} \\&=1.6 \mathrm{~A}\end{aligned}$$

Potential difference across B$ and $C$,

$$\begin{aligned}&=V_{B}-V_{C} \\&=1.6 \times 20 \\&=32 \mathrm{~V}\end{aligned}$$

As,

$$V_{c}=0$$

So,

$V_{B}=+32 \mathrm{~V}$

(b) Current through arm ADC,

$$\mathrm{I}^{\prime}=\frac{48}{5+11}$$

$=3 \mathrm{~A}$

Potential difference across D and C,

\begin{aligned}&=V_{D}-V_{C} \\&=3 \times 11 \\&=33 \mathrm{~V}\end{aligned}$$

As,

$$V_{c}=0$$

So,

$$V_{D}=33 \mathrm{~V}$$

(c) As $V_{D} > V_{C I}$

So direction of current will be from D$ to $B$.

(d) The bridge will be balanced when :

$$\begin{aligned}&\frac{10}{R}=\frac{5}{11} \\&\text { or } R=\frac{10 \times 11}{5} \\&=22 \Omega .\end{aligned}$$

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