Question

The four arms ABCD of a Wheatstones network have the following resistances: AB = 2 Ω, BC=4 Ω, CD= 4 Ω and DA= 8 Ω. A galvanometer of resistance 10 Ω is connected between B and D. Find the current through the galvanometer, when the potential difference between A and C is 5 V.

Answer 1

The current through the galvanometer is 1.071 A.

Explanation:

Given that,

AB = 2 Ω

BC=4Ω

CD = 4 Ω

DA = 8Ω

Galvanometer resistance = 10 Ω

Potential difference between A and C = 5 V

We need to calculate the potential difference

Using Kirchhoff's current law at node B

$\dfrac{v_{1}-5}{2}+\dfrac{v_{2}-v_{1}}{10}-\dfrac{v_{1}}{4}=0$

$3v_{1}+2v_{2}=50$....(I)

Using Kirchhoff's law at node D

$\dfrac{v_{2}-5}{8}+\dfrac{v_{1}-v_{2}}{10}-\dfrac{v_{2}}{4}=0$

$-9v_{2}+4v_{1}=25$.....(II)

From equation (I) and (II)

$35v_{2}=125$

$v_{2}=\dfrac{125}{35}$

$v_{2}=3.57\ V$

Put the value of v₂ in to the equation (I)

$3v_{1}+2\times3.57=50$

$v_{1}=\dfrac{50-2\times3.57}{3}$

$v_{1}=14.28\ V$

We need to calculate the current through the galvanometer

Using formula of ohm's law

$I=\dfrac{V_{2}-v_{1}}{10}$

Put the value into the formula

$I=\dfrac{14.28-3.57}{10}$

$I=1.071\ A$

Hence, The current through the galvanometer is 1.071 A.

Learn more :

Topic : Kirchhoff's law

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Answer 2

The current through the galvanometer, when the potential difference between A and C is 5 V is 1.071 ampere.

Explanation:

On applying Kirchoff's current law at node B, we get,

(V₁ - 5)/2 + (V₂ - V₁)/10 - V₁/4 = 0 → (equation 1)

On applying Kirchoff's current law at node D, we get,

(V₂ - 5)/8 + (V₁ - V₂)/10 - V₂/4 = 0 → (equation 2)

On solving equation (1) and (2), we get,

V₁ = 14.28 volt

V₂ = 3.57 volt

Now, the current is given by the formula:

I = V/R

Where,

V = Potential difference

R = Resistance

Now, the current is:

I = (14.28 - 3.57)/10

I = 10.71/10

∴ I = 1.071 A