Question

A Wheatstone bridge ABCD has the following details; AB = 1000 Q; BC = 100 Q; CD = 450 Q; DA = 5000 Q. A galvanometer of resistance 500 Q is connected between B and D. A 4.5-volt battery of negligible resistance is connected between A and C with A positive. Find the magnitude and direction of galvanometer current​

Answer 1

Explanation:

The four arms of a wheatstone bridge ABCD have the following resistances. AB = 1000, BC = 150 resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Answer 2

The Main Answer is: The current flowing in the galvanometer is 37.4 μA from B to D.

Given: Wheatstone bridge ABCD

           AB = 1000 Ω; BC = 100 Ω; CD = 450 Ω; DA = 5000 Ω

           4.5 V battery between A and C

           Galvanometer between B and D of resistance = 500 Ω

To Find: Magnitude and direction of current in the galvanometer.

Solution:

Let current flowing in AB = $i_{1}$

Let current flowing in AD = $i_{2}$

Let current flowing through galvanometer = $i_{g}$ from B to D (assume)

Therefore, the current flowing in BC = $i_{1} - i_{g}$

The current flowing in DC = $i_{2} + i_{g}$

Kirchoff's Voltage Law (KVL) states that voltage in a closed-loop adds up to zero.

Using KVL-

In loop ABDA-

-1000i1$i_{1}$ - 500$i_{g}$ + 5000$i_{2}$ = 0

-2$i_{1}$ - $i_{g}$ + 10$i_{2}$ = 0

$i_{g}$ = 10$i_{2}$ - 2$i_{1}$ ------(1)

In loop BCDB-

-100($i_{1}$ - $i_{g}$) + 450($i_{2}$ + $i_{g}$) + 500$i_{g}$ = 0

-100$i_{1}$ + 450$i_{2}$ + 1050$i_{g}$ = 0

-2$i_{1}$ + 9$i_{2}$ + 21$i_{g}$ = 0

$i_{2}$ = 1/9(2$i_{1}$ - 21$i_{g}$) --------(2)

In loop EABCFE-

4.5 - 1000$i_{1}$ - 100($i_{1}$ - $i_{g}$) = 0

4.5 = 1100$i_{1}$ - 100$i_{g}$ ---------(3)

Put (2) in (1)

$i_{g}$ = 10/9(2$i_{1}$ - 21$i_{g}$) - 2$i_{1}$

9$i_{g}$ = 20$i_{1}$ - 210$i_{g}$ - 18$i_{1}$

$i_{1}$ = 219$i_{g}$/2 ------(4)

Put (4) in (3)

1100(219/2)$i_{g}$ - 100$i_{g}$ = 4.5

240900$i_{g}$ - 200$i_{g}$ = 9

240700$i_{g}$ = 9

$i_{g}$ = 37.4 × $10^{-6}$ A

$i_{g}$ = 37.4 μA

Since this value came positive, it means that our assumption of the direction of current is correct, i.e. from B to D.

Therefore, the current through the galvanometer is 37.4 μA from B to D.

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