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

Explanation

Given emf of each cell= 2.0V
Total emf of six cells `= 2xx 6 = 12V`
Total internal resistance= `6 xx 0.015Omega = 0.090Omega`
Load resistance = `8.5Omega`
Current in the circuit I= `(E_("eq"))/(R+r_("eq"))`
i.e., `I=12/(8.5+0.09)`
i.e., `I=12/(8.59)=1.4A`
The terminal voltage ofthe battery= V = IR
i.e., `V = 1.4 xx 8.5V`
i.e., V=11.9V