"Two resistors having resistances of 5.0 Ω and 9.0 Ω are connected in parallel. A 4.0-Ω resistor is then connected in series with the parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination. What is the current through the 9.0-Ω resistor?"

Explanation:

Given two resistor 5.0 Ω and 9.0 Ω connected on parallel

Let R1 = 5Ω, R2 = 9Ω

Equivalent of parallel connection

1/Req = 1/R1 + 1/R2

1/Req = 1/5 + 1/9

1/Req = 14/45

Then, Req(12) = 45/14 = 3.21 Ω

Req(12) = 3.21 Ω

Another 4.0 Ω resistor is connected in series to their equivalent

Let R3 = 4Ω

Total equivalent resistance

Req = R1 + R3

Req = Req(12) + R3

Req = 3.214 + 4

Req = 7.214 Ω

A voltage of 6V is connect across the terminal series connection

Then, since equivalent resistance is 7.214Ω, we can apply ohms law, to know the current that flows in the circuit

V = iR

V = Ieq × Req

6 = Ieq × 7.214

Ieq, = 6/7.214

Ieq = 0.832 A

This is the total current that flows in the circuit, since R3 is in series with the combination of R2 and R1, then, the same current flows in R3 and the combination of R1 and R2

So, R1 and R2, will have to share the currents Ieq,

So, using current divided rule to find currents in R2

I2 = R1/(R1+R2) × Ieq

I2 = 5/(5+9) × 0.832

I2 = 0.296

I2 ≈ 0.3 Amps

Check attachment for circuit diagram and another method

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