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General Physics

Four cells each of internal resistance 0.8 ohm and emf 1.4V, are connected (i) in series (ii) in parallel. The terminals of the battery are joined to the lamp of resistance 10 ohm. Find the current through the lamp and each cell in both the cases.

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9 Years agoGrade
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1 Answer

Profile image of Vikas TU
9 Years ago
Dear Student,
In case of series connection
Effective resistance = 4* 0.8 = 3.2 ohm
Then Current (I) = V/R+r
                            = 1.4 /13.2
                            = 0.1 A
In case of Parallel => effective internal resistance = 4/R = 4/0.8 = 5 ohm
Total Resistance = 15 ohm
Then current  = voltage / Resistance = 1.4 /15
                                                           = 0.09 A.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)