Question

The four arm of wheat stone bridge have the resistance 100 ohm 10 ohm 5ohm 60ohm .A galvanometer of 15ohm calculate the current through the galvanometer when a potential. Difference of 10 v is maintained acroos ac

Answer 1

Answer:

The four arms of a Wheatstone bridge  have the following resistances:

AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω.

A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Explanation:

Considering the mesh ABDA, we have

100I1 + 15Ig – 60I2 = 0

or 20I1 + 3Ig – 12I2 = 0  ——(a)

Considering the mesh BCDB, we have

10 (I1 – Ig) – 5 (I2 + Ig) – 15Ig = 0

10I1 – 30Ig – 5I2 = 0

2I1 – 6Ig – I2 = 0 ——(b)

Considering the mesh ADCEA,

60I2 + 5 (I2 + Ig) = 10

65I2 + 5Ig = 10

13I2 + Ig = 2 ——(c)

Multiplying Eq. (b) by 10

20I1 – 60Ig – 10I2 = 0 ——(d)

From Eqs. (d) and (a) we have

63Ig – 2I2 = 0

I2 = 31.5Ig ——(e)

Substituting the value of I2 into Eq. [(c)], we get

13 (31.5Ig ) + Ig = 2

410.5 Ig = 2

Ig = 4.87 mA.

Answer 2

Answer:

The current through the galvanometer is 4.87mA

Explanation:

Given, the resistances of a Wheatstone bridge  

                       $AB = 100\Omega , BC = 10\Omega , CD = 5\Omega, DA = 60\Omega$

          resistance of galvanometer,  $R= 15\Omega$

          potential difference, $V= 10 V$

Applying Kirchchoff's Voltage law to the mesh ABDA,

                   $100I_{1} + 15I_{G} -60I_{2} = 0\implies 20I_{1} + 3I_{G} -12I_{2} = 0$         ...(1)

Applying Kirchchoff's Voltage law to the mesh BCDB,

                   $10\big(I_{1} -I_{G}\big) -15I_{G}-5\big(I_{2} +I_{G}\big) = 0$  

                             $\implies 10I_{1} -30I_{G} -5I_{2} = 0\implies \big2(10I_{1} -30I_{G} -5I_{2}\big)= 0$

                             $\implies 20I_{1} -60I_{G} -10I_{2}= 0$                                         ...(2)

Substracting equation (2) from (1),

                            $\implies 63I_{G} -2I_{2}= 0\implies2I_{2} =63I_{G}$

                                                              $\implies I_{2} =31.5I_{G}$                           ...(3)

Applying Kirchchoff's Voltage law to the mesh ADCEA,

                  $60I_{2} +5\big(I_{2} +I_{G}\big) = 10$

                       $\implies65I_{2} +5I_{G}\big = 10\implies13I_{2} +I_{G}\big = 2$

                       $\implies I_{G}\big = 2-13I_{2}$

Using equation (3),

                    $I_{G} = 2-13\big(31.5I_{G}\big)\implies I_{G} = 2-409.5I_{G}$

     $\implies 410.5I_{G} = 2\implies I_{G} = \frac{4}{410.5}$

                                              $=0.00487A=4.87mA$

Hence, the current through the galvanometer is 4.87mA