Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

Solution

Given,
Emf of secondary cells, $E = 2.0 V$
Internal resistance of cell, r= 0.015 $Ω$
Number of secondary cells, n=6
External resistance, R = 8.5 $Ω$

Current drawn from the supply, I = $I = \frac{nE}{R + nr}$
   $= \frac{6 \times 2.0}{8.5 + 6 \times 0.015} = 1.4 A$

And, terminal voltage of the supply is, V=IR
   $= 1.4 \times 8.5 = 11.9 V$

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Current Electricity

Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

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