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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?
one month ago

Arun
23523 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
one month ago
Vikas TU
10078 Points

Dear student
The Resistance value is incorrect
Applying Kirchoff voltage rule we get
net emf =     (120-12)=117 V , net resistance =( R+r)   and net current =3A
​now 117=i xRnet
117/3 =r+R
39-0.5 =R=38.5 Ω
Hope this will help
one month ago
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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