Question

Three resistors having resistances of LaTeX: 4.0\;\Omega4.0 Ω, LaTeX: 6.0\;\Omega6.0 Ω, and LaTeX: 10.0\;\Omega10.0 Ω are connected in parallel. If the combination is connected in series with an ideal 12-V battery and a LaTeX: 2.0\;\Omega2.0 Ω resistor, what is the current through the ten-ohm resistor?

Answer 1

Step-by-step explanation:

Given that three resistor of resistance

R1 = 4 Ω

R2 = 6 Ω

R3 = 10 Ω

The resistor are connected in parallel, the equivalent resistance is

1 / Req = 1 / R1 + 1 / R2 + 1 / R3

Rearranging this we have

Req = R1•R2•R3 / (R2•R3 + R1•R3 + R1•R2)

Req = 4 × 6 × 10 / (6×10 + 4 × 10 + 4 × 6)

Req = 240 / (60 + 40 + 24)

Req = 240 / 124

Req = 1.94 Ω

The parallel connection is connected in series with a battery and a resistor.

Resistor resistance is

r = 2 Ω

Supply voltage of the battery is 12V

V = 12V

Let find the current flowing in the circuit.

V = I(Req + r)

I = V / (Req + r)

I = 12 / (1.94 + 2)

I = 12 / 3.94

I = 3.05A

So, this is the current flowing in the circuit and It is the same current that will be shared by the parallel resistance.

We can calculated the voltage across the parallel resistance

From ohms las

V = iR

V = I × Req

V = 3.05 × 1.94

V = 5.92 V.

Then, this the voltage across each parallel resistor.

Then, to know the current in the 10ohms resistance

V = iR

I = V / R3.

I = 5.92 / 10

I = 0.592 A.

The current in the 10 ohms resistor is 0.59A