Question

Find the current drawn from the battery by networks
four resistors shown in the figure.
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Answer 1

Answer:

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Answer 2

Answer:

hey mate here is your is answer

Explanation:We know that AD is the shortest way but ABCD is connected in series

Equivalent Resistance =R1+R2+R3

=(10+10+10)ohm

=30ohm

Since BC and AD is connected in parallel

1/Req=1/R1 + 1/R2

=1/30 + 1/10

=3+1/30

=4/30

So, Req=30/4 ohm

Now,Current drawn from the battery

=> I = V/R

=> I = 30/4 ohm/3v

=> I = 30/4*3

So, I = 10/4 A