Current Electricity 4 Question 6

6. The galvanometer deflection, when key K1 is closed but K2 is open equals θ0 (see figure). On closing K2 also and adjusting R2 to 5Ω, the deflection in galvanometer becomes θ05. The resistance of the galvanometer is given by (neglect the internal resistance of battery) :

(Main 2019, 12 Jan I)

(a) 22Ω

(b) 5Ω

(c) 25Ω

(d) 12Ω

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Solution:

  1. For a galvanometer, igθ or ig=Cθ

where, Ig= current through coil of galvanometer,

θ = angle of deflection of coil

and C is the constant of proportionality.

Now, K1 is closed and K2 is opened, circuit resistance is

Req=R1+Rg

where, Rg= galvanometer resistance.

Hence, galvanometer current is

i.e.,

ig=VR1+Rg

where, V= supply voltage.

As deflection is given θ0, we have

ig=VR1+Rg=Cθ0

When both keys K1 and K2 are closed, circuit resistance.

Req=R1+R2×RgR2+Rg

Current through galvanometer will be

ig2=VR1+R2RgR2+Rg×R2(R2+Rg)=VR2R1R2+R1Rg+R2Rg=Cθ05

Now, dividing Eq. (i) by Eq. (ii), we get

R1R2+R1Rg+R2RgR2(R1+Rg)=5

Substituting R1=220Ω and R2=5Ω,

we get

1100+220Rg+5Rg5(220+Rg)=51100+225Rg=5500+25Rg200Rg=4400

Rg=4400200=22ΩRg=22Ω



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