SATHEE: Current Electricity - Result Question 102

Current Electricity - Result Question 102

####107. Three resistances $P, Q, R$ each of $2 Ω$ and an unknown resistance $S$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 Ω$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$ ?

(a) $3 Ω$

(b) $6 Ω$

(d) $2 Ω$

[2007]

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Answer:

Correct Answer: 107. (a)

Solution:

(a) A balanced wheatstone bridge simply requires

$\frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{2}{2} = \frac{2}{S}$

Therefore, $S$ should be $2 Ω$ .

A resistance of $6 Ω$ is connected in parallel.

In parallel combination,

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{1}{2} = \frac{1}{6} + \frac{1}{S} \Rightarrow S = 3 Ω$

If the resistance is connected in parallel to the right gap resistor in the metre bridge, then balancing length increases and hence jockey moves towards right.

Current Electricity - Result Question 102

Answer:

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Current Electricity - Result Question 102

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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