Current Electricity 4 Question 5

5. An ideal battery of 4V and resistance R are connected in series in the primary circuit of a potentiometer of length 1m and resistance 5Ω. The value of R to give a potential difference of 5mV across 10cm of potentiometer wire is

(2019 Main, 12 Jan I)

(a) 395Ω

(b) 495Ω

(c) 490Ω

(d) 480Ω

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Solution:

  1. Given, potential difference of 5mV is across 10m length of potentiometer wire. So potential drop per unit length is

=5×10310×102=5×102Vm

Hence, potential drop across 1m length of potentiometer wire is

VAB=5×102Vm×1=5×102V

Now, potential drop that must occurs across resistance R is

VR=45×102=395100V

Now, circuit current is

i=VRtotal =4R+5

Hence, for resistance R, using VR=iR, we get

395100=4R+5×R395(R+5)=400R395×5=(400395)R

R=395Ω



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