Question

Three resistors connected in parallel have individual values of 4.0, 6.0, and 10.0 Ω, respectively. If this combination is connected in series with a 12-V battery and a 2.0- Ω resistor, what is the current in the 10.0- Ω resistor?

Answer 1

Answer:

The current in the 10.0- Ω resistor is 1.16A

Explanation:

When resistors are connected in series, they are have the same current.

When they are in parallel, their current is divided.

The equivalent resistance of parallel resistors R1 and R2 is:

Req = R1*R2/(R1 + R2)

Initially, we use Ohm's Rule to solve this question.

$V = Ri$

In which V is the Voltage, R is the resistence and i is the current.

Among two resistors, R1 and R2, in parralel, we have that

iR1 = i(0)R2/(R1 + R2)

In which i(0) is the current entering.

Three resistors connected in parallel have individual values of 4.0, 6.0, and 10.0 Ω, respectively. In series with a 12-V battery and a 2.0- Ω resistor.

The current entering the parallel is the same current as it is for the 2.0- Ω resistor, with 12 V.

It is

$V = Ri$

$12 = 2i$

$i = 6A$

This means that i(0) = 6A.

4.0, 6.0, and 10.0 Ω in parallel

We want the current in the 10.0 Ω resistor.

The current divisor is not applicable with 3 resistors. However, we can calculate the equivalent resistance for the resistors of 4 and 6, and then the current divisor with the 10. So

Req = 4*6/(4+6) = 2.4

iR10 = 6*2.4/(10 + 2.4) = 1.16

The current in the 10.0- Ω resistor is 1.16A