Question

Please give the answer fast​

Answer 1

Answer:

Ist Rp =1/ R1 + 1/R2 + 1/R3

Rp = 1/1 + 1/1 + 1/1

Rp = 3....

In second case,

2nd Rp = 1/2+1/2+1/2

= 3/2 = 1.5

Now,

Total resistance = Rs = 1st Rp + 2nd Rp

Rs = 3 +1.5= 4.5

By applying ohm's law we have,

V = IR

I = V/R

I = 10/4.5 = 2.22 Ampere

Hence, The total resistance = 4.5 Ω and total current = 2.22 Ampere

Explanation:

Hope it helps!!

Answer 2

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,

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

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