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A storage battery of emf 12V and internal resistance 0.5 ohm, is to be charged by a 120V D.C supply of negligible internal resistance. What resistance is required in the circuit for the charging current to be 3A? Calculate: i)How much chemical energy from the dc source is converted to electrical energy? ii)How much electrical energy is converted to chemical energy in the storage battery?

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8 months ago

Arun
24738 Points
```							 By applying Kirchoff's voltage law , we get,   3R + 12 + (3×0.5) = 120  From abov eqn., we get R = 35.5 Ω If t is charging time in seconds, electrical energy converted into chemical energy in storage battery = ( 12 × 3 × t ) J = 36t J If t is charging time in seconds, chemical energy converted into electrical energy in dc source = ( 120 × 3 × t ) J = 360t J
```
8 months ago
Vikas TU
12133 Points
```							Dear student The Resistance value is incorrect Applying Kirchoff voltage rule we getnet emf =     (120-12)=117 V , net resistance =( R+r)   and net current =3A​now 117=i xRnet117/3 =r+R39-0.5 =R=38.5 ΩHope this will help
```
8 months ago
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### Course Features

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### Course Features

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