Question

Four resistors of resistance 20Ω, 20Ω, 20Ω and 30 Ω form a Wheatstone’s network. What shunt is required across 30Ω resistor to balance the bridge? a) 10Ω b) 30Ω c) 20Ω d) 60Ω



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Answer 1

Answer:

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Four resistors of resistance 20Ω, 20Ω, 20Ω and 30 Ω form a Wheatstone’s network. What shunt is required across 30Ω resistor to balance the bridge? a) 10Ω b) 30Ω c) 20Ω d) 60Ω

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The Wheatstone’s bridge is more sensitive when all their resistances are equal or or their ratio is unity.

The shunt "s" should be connected in parallel.

$\therefore \: p = 20 \\ \\ q = 20 \\ \\ r = 20 \\ \\ s = 30 + r$

$\therefore \: \frac{ \cancel{20}}{ \cancel{20}} = \frac{(30 + r)20}{30r} \\ \\ \therefore \: 30r = (30 + r)20$

So equation is

$30r = (30 + r)20$

Now after solving it,

the simplifying part will be

$10r = 600 \\ \\ \therefore \: r = \frac{60 \cancel0}{ 1\cancel0} \\ \\ \boxed{ \green{r = 60}}$

Option d) is correct

Hope it helps you from my side

:)

Answer 2

Answer:

Explanation:

This question doesn't demand much from you except for the formula of resistors in parallel,how is a shunt resistance connected in a circuit and the balance condition of the whetstone bridge.

So,the formula for resistors in parallel :

1/R1 + 1/R2 + 1/R3...

The shunt is actually put into parallel with the galvanometer which is used in series with the circuit to measure the current strength.

The wheatstone bridge is said to be balanced if :

R1/R2 = R3/R4

Using all these basic things we can easily solve this question. Let's begin.

So, first of all put the shunt in parallel to any of the resistance, let's take it in parallel with 30 Ω resistor (you can take any resistor,the answer will be the same).

So, you are required to first of all calculate the effective resistance of the shunt and the 30 Ω resistors put in parallel. Let s denote the shunt resistance.

1/Req = 1/s + 1/30

1/Req = 30 + s / 30s

Req = 30s / 30 + s -----> (1)

Now,use the balance condition of the wheatstone bridge. The R4 resistor will now become Req because we have take in account the shunt resistance too.

R1/R2 = R3/Req

Put the values,

20/20 = 20/ (30s/ 30 + s)

1 = 20 × (30 + s)/ 30s

30s = 600 + 20s

30s - 20s = 600

10s = 600

s = 600/10

s = 60 Ω

∴ 60 Ω shunt is required across 30 Ω to balance the bridge.

Now this can also be done in another way.

First let's apply the balance condition of wheatstone bridge keeping the shunt in parallel, let's call the combined resistance of shunt and 30 Ω "s".

R1/R2 = R3/Rcombined

20/20 = 20/s

1 = 20/s

s = 20

And we know that the shunt is connected in parallel with the 30 Ω resistor,so applying the formula for parallel combination of resistors, let's call the unknown shunt required to balance the bridge as x.

1/s = 1/x + 1/30

1/20 - 1/30 = 1/x

30 - 20 / 600 = 1/x

10/600 = 1/x

1/60 = 1/x

Cross multiply,

x = 60

∴ Option (d) 60 Ω is your answer to this question.

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