Given circuit below
Resistance between
B and
C on the right hand side of the circuit
===5Ω because
1Ω+4Ω+5Ω=10Ω are in parallel with
10Ω.
Resistance between
A and
D on the right hand side of the circuit
===5Ω because
5Ω (equivalent resistance of
1Ω,4Ω,5Ω and
10Ω ) is in series with
3Ω and
2Ω.
Hence, resistance between
A and
D will be
5Ω.
Equivalent resistance of the circuit,
R=5+2+5=12Ω Current drawn from the battery,
I==1A At the junction
A, the current of
1A is divided equally between the
10Ω resistance and the remaining circuit of the
10Ω resistance.
At the junction
B, the current of
0.5A is divided equally between the
10Ω resistor and remaining circuit of resistance
10Ω.
∴ Current through the
4Omega resistor
=0.25A.