Question 34: Current Electricity

Question

Ten cells, each having internal resistance $1 \Omega$ and emf 1.5 V are connected in series. But unknowingly 3 cells are connected wrongly in series. The effective internal resistance and emf of the series combination are respectively:

Accepted Answer

HereÃ¢ÂÂs how it works out:
Internal resistances always add in series, regardless of polarity of the cells. So
$$r_{m eq}=10	imes1\;\Omega=10\;\Omega.$$
For the emfs, 7 cells contribute +1.5 V each and 3 are reversed (Ã¢ÂÂ1.5 V each):
$$E_{m eq}=(7-3)	imes1.5\;{m V}=4	imes1.5\;{m V}=6\;{m V}.$$
So the effective internal resistance and emf are
10 ÃÂ©
and
6 V
, respectively. (Option C)

Answer

$0.1 \Omega$ and 1.5 V

Answer

$6 \Omega$ and 10 V

Answer

$10 \Omega$ and 6 V

Answer

$10 \Omega$ and 15 V

Current Electricity

Ten cells, each having internal resistance 1 \Omega and emf 1.5 V are connected in series. But unknowingly 3 cells are connected wrongly in series. The effective internal resistance and emf of the series combination are respectively:

Choose the correct answer: