current-electricity Question 54

Question: Q. 7. In the given circuit, a metre bridge is shown in the balanced state. The metre bridge vire has a resistance of 1Ωcm1. Calculate the unknown resistance X and the current drawn from the battery of a negligible internal resistance if the magnitude of Y is 6Ω. If at the balancing point, we interchange the position of galvanometer and the cell, how it will affect the position of the galvanometer?

A&E [CBSE SQP, 2016]

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Solution:

Ans. } \quad XY=4060=23 X=4Ω. $$

4Ω and 6Ω are in series, =10Ω

40Ω and 60Ω are in series, =100Ω

10Ω and 100Ω are in parallel, =1000110Ω=9.09Ω1

There will be no change in the balancing length. 1/2 Formulae for series and parallel :

For parallel

R=R1R2R1+R2

For series

R=R1+R2

[CBSE Marking Scheme 2016]

[AI Q. 8.In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emf E1 and E2 connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) E1/E2 and (ii) position of null point for the cell E1.

How is the sensitivity of a potentiometer increased?

A [Delhi I, II, III 2012]

Ans. (i) E1+E2=300 K

( K is potential gradient in volt /cm )

E1E2=73

(ii)

E1+E2=300 K E1+37E1=300 K

E1=210 K

Therefore, balancing length for cell E1 is 210 cm.1/22 (Award this 1/2 mark even if the student writes E1=240 K)

(Award full marks for any other correct method)

By decreasing potential gradient, the sensitivity of a potentiometer is increased.

[Or through increasing length, reducing potential drop across wire, increasing resistance input in series with the main cell etc.]

[CBSE Marking Scheme 2012]



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