Question

How many minimum number of section will be required for potential drop to fall less than 1% if $R=R^{\prime }$?

• A. 3
• B. 5
• C. 6
• D. 4

Answer 1

Answer: (D)

4

$\frac{R}{R_1}=\frac{6.4\times {10}^3}{800\times {10}^6}=\frac{6.4}{8\times {10}^5}=0.8\times {10}^{-5}$

$=\frac{8}{10}\times {10}^{-5}=8\times {10}^{-6}$

If, $R=R^1$ then minimum number of sections $=4.$