Question

Find equivalent resistance in the above circuit. Please explain the calculation. I will mark you brainliest. ​

Answer 1

Answer:

marke as the brainliest

Answer 2

Answer:

$\huge \boxed{ \bold{\frac{3}{2} \Omega }}$

Explanation:

Here , In this circuit

1ohm and 1ohm are in parallel combination .so equivalent of these two is :

$\leadsto \: \frac{1}{r } = \frac{1}{1} + \frac{1}{1} \\ \\ \leadsto \: \frac{1}{r} = \frac{2}{1} \\ \\ \leadsto \: r = \frac{1}{2} \Omega$

Now this r is in series with another 1 ohm

so, R equivalent is

$\leadsto \: r = 1 + \frac{1}{2} \\ \\ \leadsto \: r = \frac{3}{2} \Omega$

Solution is in attachment