Question

The effective resistance of the given circuit is:

A.1Ω
B.1.5Ω
C.2Ω
D.3Ω

Answer 1

In order to solve this numerical we need to know how to solve the effective resistance. In order to find the effective resistance we consider the resistors which is connecting parallel to each other then we find the effective resistance of the circuit using parallel connection.

Complete step by step answer:
By using the series combination formula,
Between A and C, both the resistances connected in series
$\ {R_s} = {R_1} + {R_2} \\\ \implies {R_s} = 2 + 1 \\\ \implies {R_s} = 3\Omega \\\ \$
By using the parallel combination formula,
Between B and C, both the resistances connected in parallel,
$\ \dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} \\\ \implies \dfrac{1}{{{R_p}}} = \dfrac{1}{2} + \dfrac{1}{2} \\\ \implies \dfrac{1}{{{R_p}}} = 2\left( {\dfrac{1}{2}} \right) \\\ \implies {R_p} = 1\Omega \\\ \$
Now we calculate the effective resistance of the circuit
$\ \dfrac{1}{{{R_{{p_{total}}}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} \\\ \implies \dfrac{1}{{{R_p}}} = \dfrac{1}{3} + \dfrac{1}{3} \\\ \implies \dfrac{1}{{{R_p}}} = \left( {\dfrac{2}{3}} \right) \\\ \implies {R_p} = \left( {\dfrac{3}{2}} \right) \\\ \implies {R_p} = 1.5\Omega \\\ \$

Note:
Commonly the Parallel circuit connection is more in use. Different electrical appliances we use in our daily life are usually connected in parallel such that electrical appliances can work independently. We can have control over the individual electrical appliances such that we need to be wired in parallel circuits.