Question

4) Find current drawn from battery
by network of four resistors.
find no resistor In series and in parallel
​

Answer 1

Answer:

HOPE IT HELPS UH !

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