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Earlier discussion
1 older comment kept for context.
Oluwaseun Adeyemi
The sum of the currents I1 and I2 in the figure above is
The resistor 10Ω is in parallel with 5Ω and 10Ω. Therefore, the same potential difference will flow through them.
Note that 5Ω and 10Ω is in series
V = 12V
R = 10Ω
V = IR
12 = 10I2
I2 = 12/10 = 1.2A
5Ω and 10Ω in series
Reff = 5Ω + 10Ω = 15Ω
V = 12V
V = IR
12 = 15I1
I1 = 12/15 = 0.8A
I = I1 + I2 = 1.2A + 0.8A = 2A