Question

#Quality answer needed
A circuit consists of 1 Omega wire in series with a parallel arrangement of 6 Omega and 3 Omega wires. Calculate total resistance of the circuit.​

Answer 1

Answer:

Explanation:

First, let us know the setup of the question.

For series :- $\sf{R_{eq} = R_1 + R_2}$

Where R1, R-eq and R2 are the resistances in the wires respectively.

Now, for Parallel combination :-

$\sf{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}$

Now, in this question, we need to deal with Parallel arrangement first, to evade complexity.

$\sf{\dfrac{1}{R_{eq}} = \dfrac{1}{6} + \dfrac{1}{3}}$

$\sf{\dfrac{1}{R_{eq}} = \dfrac{2+1}{6}}$

$\sf{\dfrac{1}{R_{eq}} = \dfrac{3}{6}}$

$\sf{\dfrac{1}{R_{eq}} = \dfrac{1}{2}}$

$\sf{R_{eq} = 2 \Omega}$

Now, we can apply the formula for series combination, which will be

$\sf{R_{eq} = R_1 + R_2}$

$\sf{R_{eq} = 2 + 1}$ (One from the question, other we determined previously)

$\boxed{\sf{R_{eq} = 3 \Omega}}$ is the answer.