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Question: A torch battery consisting of two cells of 1.45 volts and an internal resistance\(0.15 \Omega\), eac...

A torch battery consisting of two cells of 1.45 volts and an internal resistance0.15Ω0.15 \Omega, each cell sending currents through the filament of the lamps having resistance 1.5ohms. The value of current will be

A

16.11 amp

B

1.611 amp

C

0.1611 amp

D

2.6 amp

Answer

1.611 amp

Explanation

Solution

Here two cells are in series.

Therefore total emf = 2E.

Total resistance = R + 2r

∴ i=2ER+2r=2×1.451.5+2×0.15=2.91.8=2918=1.611ampi = \frac { 2 E } { R + 2 r } = \frac { 2 \times 1.45 } { 1.5 + 2 \times 0.15 } = \frac { 2.9 } { 1.8 } = \frac { 29 } { 18 } = 1.611 \mathrm { amp }