Question

Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network. What shunt is required across 15 Ω resistor to balance the bridge
(a) 10 Ω (b) 15 Ω (c) 20 Ω (d) 30 Ω​

Answer 1

Answer:

option (d) 30 Ω

Explanation:

Given :

Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network.

To find :

the shunt required across 15 Ω resistor to balance the bridge.

Solution :

The bridge is said to be in balanced condition when no current passes through the Galvanometer.

The condition for the balanced bridge is

$\boxed{ \bf \: \dfrac{P}{Q} = \dfrac{R}{S}}$

here,

P = 10 Ω

Q = 10 Ω

R = 10 Ω

Let Rₛ be the shunt resistance.

We know,

When the resistors of resistances R₁, R₂ are connected in parallel, the equivalent resistance is given by,

R' = (R₁R₂)/(R₁ + R₂)

So,

S = (15Rₛ)/(15 + Rₛ)

Substitute the values,

$\longrightarrow \sf \dfrac{10}{10} = \dfrac{10}{\dfrac{15R_s}{15+R_s }} \\ \longrightarrow \sf 1 \times \dfrac{15R_s}{15+R_s} = 10$

⟶ 15Rₛ = 10(15 + Rₛ)

⟶ 15Rₛ = 150 + 10Rₛ

⟶ 15Rₛ – 10Rₛ = 150

⟶ 5Rₛ = 150

⟶ Rₛ = 150/5

⟶ Rₛ = 30 Ω

Therefore, the required shunt across 15 Ω is 30 Ω