Question

In the wheatstone bridge network shown in figure. The current passing through battery is​

Answer 1

First of all we have to find equivalent resistance of the given circuit.

• In the wheatstone bridge current doesn't pàss through the resistance which is connected in the middle part. Hence we will not consider resistance of 5Ω in the calculation of net equivalent resistance.

1. Equivalent R of ABC branch :

In this branch both resistors are connected in series.

➝ R₁ = 1 + 2

➝ R₁ = 3 Ω

2. Equivalent R of ADC branch :

In this branch also two resistors are connected in series.

➝ R₂ = 2 + 4

➝ R₂ = 6 Ω

3. Equivalent R of R₁ and R₂ : [AC]

After calculating equivalent resistance of both branches, R₁ and R₂ come in parallel.

Their equivalent R₃ will be,

➝ 1/R₃ = 1/R₁ + 1/R₂

➝ 1/R₃ = 1/3 + 1/6

➝ 1/R₃ = 3/6

➝ R₃ = 2 Ω

Finally R₃ and 1Ω come in series. Therefore net equivalent resistance of the circuit will be,

➠ R = R₃ + 1

➠ R = 2 + 1

➠ R = 3 Ω

As per ohm's law current flow in the circuit is directly proportional to the applied potential difference.

Mathematically, V = I R

• Where R denotes equivalent resistance of the circuit$.$

By substituting the known values;

➙ V = I R

➙ 6 = I × 3

➙ I = 2 A

Answer 2

Answer:

The balanced condition for Wheatstones bridge is

QP = SR

as is obvious from the given values.

No, current flows through galvanometer is zero.

Now, P and R are in series, so

Resistance,R 1

=P+R

=10+15=25Ω

Similarly, Q and S are in series, so

Resistance R 2

=R+S

=20+30=50Ω

Net resistance of the network as R

1 and R 2 are in parallel

i= RV

= 506×3

=0.36 A.