Current Electricity Charging Discharging Of Capacitors Question 143
Question: If in the circuit shown below, the internal resistance of the battery is 1.5 W and VP and VQ are the potentials at P and Q respectively, what is the potential difference between the points P and Q
[MP PET 2000]
Options:
A) Zero
B) 4 volts (VP > VQ)
C) 4 volts (VQ > VP)
D) 2.5 volts (VQ > VP)
Show Answer
Answer:
Correct Answer: D
Solution:
$ R _{eq}=\frac{5}{2}\Omega $
$ i=\frac{20}{\frac{5}{2}+1.5}=5A $ Potential difference between X and P, $ V _{X}-V _{P}=( \frac{5}{2} )\times 3=7.5V $ ….(i) $ V _{X}-V _{Q}=\frac{5}{2}\times 2=5V $ ?(ii) On solving (i) and (ii) $ V _{P}-V _{Q}=-2.5volt\text{; }V _{Q}>V _{P} $ . Short Trick : $ (V _{P}-V _{Q})=\frac{i}{2}(R _{2}-R _{1}) $
$ =\frac{5}{2}(2-3)=-,2.5 $
Therefore $ V _{Q}>V _{P} $
BETA
