Current Electricity Charging Discharging Of Capacitors Question 184
Question: The resistance of the series combination of two resistance is S. When they are joined in parallel the total resistance is P. If S = nP, then the minimum possible value of n is
[AIEEE 2004]
Options:
A) 4
B) 3
C) 2
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
If two resistances are $ R _{1} $ and $ R _{2} $ then $ S=R _{1}+R _{2} $ and $ P=\frac{R _{1}R _{2}}{(R _{1}+R _{2})} $ From given condition S = nP i.e. $ (R _{1}+R _{2})=n,( \frac{R _{1}R _{2}}{R _{1}+R _{2}} ) $
Therefore $ {{(R _{1}+R _{2})}^{2}}=nR _{1}R _{2} $
Therefore $ {{(R _{1}-R _{2})}^{2}}+4R _{1}R _{2}=nR _{1}R _{2} $ So $ n=4+\frac{{{(R _{1}-R _{2})}^{2}}}{R _{1}R _{2}}. $ Hence minimum value of n is 4.
BETA
