a resistance of the Potentiometer wire is 8 ohm and its length is 8 metre a resistance box and a 2 volt battery are connected in series with it what should be the resistance in the box if it is desire to have the potential difference of a 1 volt upon mm
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Answer:
In the picture it's given 1uV/mm. However, you've asked for 1 V/mm. I'll do both.
E = IRwire/lwire
=> E = I (8/8) = I = 2/(8+R)
If E = 1uV/mm, then
10^-6/10^-3 = 2/(8+R)
=> R = 1992 ohm = 1.992 mohm
If E = 1 V/mm, then
1/10^-3 = 2/(8+R)
=> R = -7.998 ohm which is impossible. So, E can't be 1 V/mm. If you observe carefully, for a feasible R, E needs to be lesser 2.5 mV/cm i.e 0.25 V/m i.e 0.00025 V/mm.
I hope the matter is clear.
Answer:
E = IRwire/lwire
=> E = I (8/8) = I = 2/(8+R)
If E = 1μV/mm, then
10^-6/10^-3 = 2/(8+R)
=> R = 1992 Ω
= 1.992 mΩ
Explanation:
Answer is R = 1992 Ω