Physics Question - Current in Resistor of parallel circuit - Need Help!?

The three R's in parallel are equivalent to

1/R = 1/5 + 1/12 + 1/20

1/R = 0.2 + 0.083 + 0.05 = 0.333

R = 3 Ω

add 9 in series to get 12 Ω

I = e/R = 9/12 = 0.75 amp

that produces a voltage across the 3 in parallel of

E = IR = 0.75 x 3 = 2.25 volts

and the currents in the 3 branches are

I = E/R

I = 2.25/5 = 0.45 amp

I = 2.25/12 = 0.1875 amps

I = 2.25/20 = 0.1125

(the three should add up to 0.75 amps)

PS, the circuit is drawn poorly, the junctions have to be shown clearly, see below.

The 5 Ω, a 12 Ω, and a 20 Ω resistor are in parallel with each other.

1/Req = 1/5 + 1/12 + 1/20 = ⅓

Req = 3 Ω

Total resistance = 9 + 3 = 12 Ω

B. I = 9 ÷ 12 = 0.75 amps

Let’s use the following equation to determine the voltage for the 9 Ω resistor.

V = I * R = 0.75 * 9 = 6.75 volt

To determine the voltage for each of the resistors in parallel, subtract this number from 9.

V = 9 – 6.75 = 2.25 volts

I 5 = 2.25 ÷ 5 = 0.45 amp

I 12 = 2.25 ÷ 12 = 0.1875 amp

I 20 = 2.25 ÷ 20 = 0.1125 amp

To check these currents, the sum should be 0.75 amps.

Sum 0.45 + 0.1875 + 0.1125 = 0.75

The total current is 0.75A as you say.

The currents in the branches are:

5R: 0.45A

12R: 0.1875A

20R: 0.1125A

Sanity check: 0.45 + 0.1875 + 0.1125 = 0.75

Try 1.8 but I’m not sure it’s right