If two resistors of 25Ω and 15Ω are joined together in series and then placed in parallel with a 40Ω resistor, the effective resistance of the combination is:(a) 0.1Ω (b) 10Ω (c) 20Ω (d) 40Ω

(c) 20 Ω      

Explanation

Given,

Resistance in series = 25 Ω and 15 Ω

Resistance in parallel = 40 Ω

To find = Effective resistance or Total resistance (E_T)

Solution:

We know that when resistors are connected in series.

It is calculated as ${R}_{T}={R}_{1}+{R}_{2}+{R}_{3}+{R}_{4}+........+{R}_{n}$

So,  ${R}_{T}=25+15$

${R}_{T}=40Ω$

Now, when resistors are connected in parallel.

It is calculated as $\frac{1}{{R}_{T}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\frac{1}{{R}_{4}}+........+\frac{1}{Rn}$

So, $\frac{1}{{R}_{T}}=\frac{1}{40}+\frac{1}{40}$

$\frac{1}{{R}_{T}}=\frac{1+1}{40}$

$\frac{1}{{R}_{T}}=\frac{2}{40}$

$\frac{1}{{R}_{T}}=\frac{1}{20}$

${R}_{T}=20\Omega $

 Thus, the effective resistance of the combination will be 20Ω