Question

Find the resultant resistance for the following arrangement. Find the current, when this arrangement is connected with 9V battery.

Answer 1

here , we know, two wire connecting in two resistance.
here we use of equipotential ,
wire have no resistance. so, potential of connecting point are same .
so, we joint the connecting point In one point .

if we do it then we see
resistance combined with parallel series .
we know ,
equivalent resistance(Req) of parallel series

1/Req = 1/R1 + 1/R2 + 1/R3

then,

1/Req = 1/3 + 1/3 + 1/3

1/Req = 3/3 =1

Req = 1 ohm

use ohm's law ,

V = IR
I = V/R =9/1 = 9A

hence, current passing through it 9A

Answer 2

All the resistors are in parallel combination.
If we break the circuit , then the resulting combination will appear somewhat like in the picture provided below.

We know that ,
1/R = 1/R₁ + 1/R₂ + 1/R₃.....
Here , R₁ = 3 ohm
R₂ = 3 ohm
R₃ = 3 ohm

1/R = 1/3 + 1/3 + 1/3
1/R = 3/3
R = 1 ohm

Now if 9V battery is connected then , current -->
I = V/R
I = 9/1
I = 9 A

So, resultant resistance = 1 ohm
Current flowing through the circuit = 9 A

Hope This Helps You!