Six secondary cells each of emf 3V and internal resistance 0.3Ω are joined in series with a 10Ω resistor. The potential drop across the external resistor is .
A
16V
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B
17.2V
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C
18V
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D
15.2V
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Solution
The correct option is D 15.2V We have 6 cells in the same direction. So their EMF’s and internal resistances get added up.
So find the total EMF and R of the combination.
E=6 × 3 = 18V
R=6 × 0.3 = 1.8 Ω
R’ = 10 Ω
Since all these are in series, we just have to find current. I=ER+R′
I=1.52A
So potential drop across 10 Δ is
V=IR’ = 15.2V