Question

How many times will be the total resistance of 'n' identical resistor increaseed if parallel arrangements is change into series arrangements?

Answer 1

Answer:

4 times

Explanation:

Taking the two different resistance

$x \:$

In parallel arrangement

$\frac{1}{r} = \frac{1}{r1} + \frac{1}{r2}$

$\frac{1}{n} = \frac{1}{x} + \frac{1}{x}$

$\frac{1}{n} = \: \frac{x + x}{x \times x}$

$\frac{1}{n} = \frac{2x}{x {}^{2} } \\ \\ n = \frac{x {}^{2} }{2x} = \frac{x}{2}$

In series arrangement

$r = r1 + r2$

$r = x + x = 2x$

Therefore to find how many times that resistance increase in series arrangement we divide resistance in series by resistance in parallel arrangement

$2x \div \frac{x}{2} \\ = 2x \times \frac{2}{ x} \\ = 4$