Four resistors are connected to a battery with a terminal voltage of 18.0 V. The resistors are, R1=61, R2=34, R3=90, and R4=11. R1 & R2 are in series, and combined are in parallel with R3, & the combination of R1, R2, & R3 are in series with R4. Find the current in the 34.0 Ω resistor. Answer in units of A.

Please help!

E = 18 V.

R1 = 61 Ohms
R2 = 34 Ohms
R3 = 90 Ohms
R4 = 11 Ohms
I2 = ?

Req. = R4 + (R1+R2)R3/(R1+R2+R3)
Req. = 11 + (95)90/185 = 11 + 46.22 = 57.22 Ohms.

I4 = E/Req. = 18/57.22 = 0.315A

I2 + I3 = 0.315A
I3 = 95/90 * I2 = 1.056I2
I2 + 1.056I2 = 0.315
2.056I2 = 0.315
I2 = 0.1532A

To find the current in the 34.0 Ω resistor, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R).

To begin, let's analyze the circuit step by step:

1. R1 and R2 are in series, so their total resistance (RT1) is the sum of their individual resistances:
RT1 = R1 + R2 = 61 Ω + 34 Ω = 95 Ω

2. R1 and R2 combined are in parallel with R3, so their total resistance (RT2) is given by the formula:
1/RT2 = 1/RT1 + 1/R3
Rearranging the equation, we have:
1/RT2 = 1/95 Ω + 1/90 Ω
1/RT2 = (90 + 95) / (90 * 95) Ω
1/RT2 = 185 / 8550 Ω
RT2 = 8550 Ω / 185

3. The combination of R1, R2, and R3 are in series with R4, so their total resistance (RT3) is the sum of RT2 and R4:
RT3 = RT2 + R4

Now that we have the total resistance, RT3, we can calculate the current (IT3) flowing through the combination of R1, R2, R3, and R4. Using Ohm's Law:
IT3 = V / RT3
IT3 = 18.0 V / RT3

Finally, to find the current specifically in the 34.0 Ω resistor (I34), we need to consider that R4 and the 34.0 Ω resistor are in parallel. The current (I34) flowing through both resistors will be the same:
I34 = IT3

With this information, we can now calculate the current in the 34.0 Ω resistor.