Given is the following circuit. What is the current I?

We cant tell you what to do differently without seeing your attempt. However, since

it's rather obvious that you are suffering from a severe case of the "Red A$s" ; here's how you do it..

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I = V/R = 10V/ (2 + [(12)*(4)]/(12 + 4) }Ohms = 10V/5 Ohms = 2 Amps

Since I is the total current in the circuit, V = I * R where R is the equivalent resistance of the circuit.

The 12 ohm and 4 ohm resistors are in parallel, so use 1/R = 1/12 + 1/4 to find out the equivalent of that part. Then add 2 ohm, which is in series. That gives you your total R for the circuit. Plug it into V = I * R.

(12 x 4) /(12 + 4) + 2 = equivalent resistance of 5 ohms.

I = (E/R) = 10/5, = 2A.

the resistance is 12//4 = 48/16=3+2=5 Ω>> I=E/R I= 10/5= 2A.

Both answers are correct = the equivalent resistance is 3 ohms + 2 ohms = 5 ohms

We have to determine what resistance the voltage source "sees"..

First the 2Ω and then the parallel combination of 12Ω and 4Ω = 12*4/(12+4) =48/16 = 3Ω

Thus the source sees 2+3 = 5Ω

Therefore I = 10V/5Ω = 2A

When the 2A enters the junction of the 12Ω and the 4Ω a current divider shows that: 2A*4Ω/(12Ω+4Ω) = 0.5A into the 12Ω

and 2A*12Ω/(12Ω+4Ω) = 1.5A flows into the 4Ω

Notice that 0.5A+1.5A = 2A and 0.5A*12Ω= 6V and 1.5A*4Ω = 6V

Notice that 10V - 2A*2Ω = 10-4 = 6V which is the voltage at the junction of the 12Ω and the 4Ω