Question

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Answer 1

Answer:

Given,

Voltage V = 10 V

Three 1 ohm resistance are connected in parallel

so,

$\frac{1}{R_ {p1}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \: \Omega \\ \therefore \: R_{p 1}= \frac{1}{3} \: \Omega \:$

and

theree 2 ohm resistance are connected in parallel .

So,

$\frac{1}{R_{p2}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\\therefore \: R_{p2} = \frac{2}{3} \: \Omega \:$

$\sf \: finally \: \: R_{p1} \: \: and \: \: R_{p2} \: \: are \: connected \: in \: \\ \sf \: series \\ \\ \therefore \: R_{eff} = R_{p1} + R_{p2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{3} + \frac{2}{3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{3}{3} \: \Omega \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 \: \Omega \: \\ \\ \green {\boxed{\sf \: resistance \: R_{eff} \: = 1 \: \Omega \: }}$

again,

We know that,

$V = IR_{eff} \\ \implies \: I = \frac{V }{R_{eff}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{10}{ 1 } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 10 \: A$

$\green {\boxed{\sf \: total \:current \:I\: = 10 \: A \: }}$