Current Electricity Charging Discharging Of Capacitors Question 679
Question: Two wires of the same metal have same length, but their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of thicker wire is $ 10\Omega . $ The total resistance of the combination will be
Options:
A) $ 10\Omega $
B) $ 20\Omega $
C) $ 40\Omega $
D) $ 100\Omega $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Length of each wire = $ \ell $ ; Area of thick wire $ ( A _{1} )=3A; $ Area of thin wire $ (A _{2})=A; $ and resistance of thick wire $ (R _{1})=10\Omega . $ Resistance $ ( R )=\rho \frac{\ell }{A}\propto \frac{1}{A} $ (if l is constant)
$ \
Therefore \frac{R _{1}}{R _{2}}=\frac{A _{2}}{A _{1}}=\frac{A}{3A}=\frac{1}{3} $ or, $ R _{2}=3R _{1}=3\times 10=30\Omega $ The equivalent resistance of these two resistors in series $ =R _{1}+R _{2}=30+10=40\Omega . $
BETA
