I need a little direction with this question:If someone could provide the correct formula, I think I divide each reistor by 12V but I'm not sure. To find total resistance, do I just add the reciprocals?

Three resistors connected in parallel have individual values of 4.0, 6.0 and 10.0 W, respectively. If this combination is connected in series with a 12.0 V battery and a 2.0 W resistor, what is the current in each of the resistors and what is the total resistance of the circuit?



a. 0.59 A

b. 1.0 A

c. 11.2 A

d. 16.0 A

Please irnore the multiple choice answers-they were for a different problem that I copied down

find total resistance of the parallel network. I assume you mean ohms, not watts on the reistors.

1/Rt=1/4+1/6+1/10=1/60 (15+10+6)=31/60
Rt=60/31

Now add that inseries to 2 ohm, or total resistance is 151/60
current= V/R=12(60/151) check that
now, having total current, how does it divide in the paralel branch.
the voltage across the parallel branch is Rt*It= 60/131*69/151)

so current in each resistor is that voltage divided by either 4,6, or 10 ohms

Thank you for your help-I can really use the explanation for my other problems too.

Thank you again-

I have another question-I looked closely at the explanation and when you added the three resistors together you came up with Rt=60/31. Then you added the series to the 2Ohm and said total equals 151/60-I'm lost there because 2 Ohms would equal 30/60 or when added to the inseries it would become resistance total = 60/61 so where does the 151/60 come from? Sorry if I'm just not understanding

I figured it out-thank you

To solve this problem, we need to use the formulas for resistors in series and resistors in parallel.

First, let's find the equivalent resistance of the three resistors connected in parallel. The formula for resistors in parallel is:

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

Here, R1, R2, and R3 are the individual resistances. Plugging in the given values:

1/Req = 1/4.0 + 1/6.0 + 1/10.0

To simplify this equation, we need to find the least common denominator (LCD) for the fractions. The LCD of 4, 6, and 10 is 60. Multiply each fraction by the LCD:

1/Req = 15/60 + 10/60 + 6/60

Now add the fractions:

1/Req = 31/60

To find Req, take the reciprocal of both sides:

Req = 60/31 ≈ 1.94 Ω

So the total resistance of the three resistors connected in parallel is approximately 1.94 Ω.

Now, since the combination of the three resistors and the 2.0 Ω resistor is connected in series with a 12.0 V battery, we can use Ohm's Law to find the current in the circuit. Ohm's Law states that:

V = I * R

Where V is the voltage, I is the current, and R is the resistance.

The total resistance of the circuit is the sum of the resistances of the three resistors and the 2.0 Ω resistor:

Rtotal = Req + 2.0

Rtotal = 1.94 + 2.0

Rtotal ≈ 3.94 Ω

Now we can use Ohm's Law to find the current (I):

12 = I * 3.94

Solving for I:

I = 12 / 3.94 ≈ 3.05 A

So the current in the circuit is approximately 3.05 A.

Since the three resistors are connected in parallel, the current is the same for each resistor. Therefore, the current in each of the resistors is also approximately 3.05 A.

Therefore, the correct answer is not listed among the options provided.