Question

Find the current in the following

Answer 1

Let R1=6 ohm,R2=6ohm,R3=2 ohm

Now if u see BCED are in parallel connection ,so

1/Rp=1/R1+1/R2

1/Rp=1/6+1/6

1/Rp=6+6/36

1/Rp=12/36

1/Rp=1/3

Rp=3ohm

Now if u see Rp and R3 are in series connection,

So,Rs=Rp+R3

Rs=3+2

Rs=5ohm

Total resistance in the circuit(Rs)=5ohm

Potential difference(V)=15V

From ohm's law,

Current(I)=V/Rs

=15/5

=3A

Therefore,current in the given circuit=3A

Hope it helps you