Current Electricity Charging Discharging Of Capacitors Question 145
Question: Two cells of equal e.m.f. and of internal resistances $ r _{1} $ and $ r _{2}(r _{1}>r _{2}) $ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be
[MP PET 1985; KCET 2005; Kerala PMT 2005]
Options:
A) $ r _{1}+r _{2} $
B) $ r _{1}-r _{2} $
C) $ \frac{r _{1}+r _{2}}{2} $
D) $ \frac{r _{1}-r _{2}}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let the voltage across any one cell is V, then $ V=E-ir=E-r _{1},( \frac{2E}{r _{1}+r _{2}+R} ) $ But V = 0
Therefore $ E-\frac{2Er _{1}}{r _{1}+r _{2}+R}=0 $
Therefore $ r _{1}+r _{2}+R=2r _{1} $
Therefore $ R=r _{1}-r _{2} $
BETA
