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`        A potentiometer wire of length 10 m and resistance 10 Ω per metre is connected in series with a resistance box and a 2 volt battery. If a potential difference of 100mV is balanced across the whole length of potentiometer wire, then the resistance introduced in the resistance will beanswer is 190ohms (im getting my answer as 1900 ohms pls help)`
one year ago

Sudarshan
13 Points
```							       For potentiometer,        I= V÷R                 I = 100 × 10^-3 ÷ 100                 I = 10^-3        Now,                 I = EMF ÷ total resistance          10^-3= 2 ÷( R + 100)                 R=1900 Ohm
```
one year ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions