Three 9W -3 V bulbs is to be lighted by a suppy of 12 Volt.For this,a wire is to be connected in series with it.Th resistance of the wire should be:A) 1
B) 2
C) 3
D) 4

Answers

Answer 1
Answer: 9W 3V bulb need current flow
I = P/V = 9/3 = 3A

connected serially, voltage drop on the wire will be = Vsupply - 3(Vbulb) = 12 - 3(3) = 3V.

resistance of the wire ,with I=3 and V=3, = V/I = 3/3 = 1 Ohm.
Answer 2
Answer: In order for a 3V bulb to dissipate 9W, the current through it must be 3A.
R = E/I = 3v/3A .  The resistance of such a bulb is 1 ohm.

There are two correct answers to the question, since the question itself
is ambiguous and may be understood in two different ways (at least) ...
series and parallel.

If the three bulbs are in series and the mystery wire will be placed
in series with them, then . . . . .

-- The current everywhere in the series circuit is 3 A .
-- The resistance across the supply is  R = E/I = 12v/3A  =  4 ohms.
-- 3 of the total 4 ohms is in the bulbs.
-- The additional wire must provide the additional 1 ohm .

If the three bulbs are in parallel and the mystery wire will be placed
in series with them, then . . . . .

-- The three bulbs in parallel form an effective resistance of 1/3 ohm.
-- The voltage across the parallel section of the circuit is 3v.
-- The additional wire must drop 9 volts across itself. 
-- The total current through the 3 branches of the parallel section,
    as well as through the additional piece of wire in series with it, is 9 A.
--  The resistance of the wire must be  R = E/I  =  9v/9A = 1 ohm.

Verrry interesting ! 
Whether the bulbs are in series or in parallel,
the same 1-ohm piece of wire does the job.

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Determine the potential difference between two charged parallel plates that are 0.20 cm apart and have an electric field strength of 6.0 V/cm.

Answers

V=Ed=0.20* 6=1.20
V=E d
d=.20 cm
E=6 V/cm
V=.20×6
  =1.2 V

A series circuit contains two devices, one with a resistance of 10 ohms and one with a resistance of 4 ohms. If the generator supplies a voltage of 42 V, what is the magnitude of the current? A. 2 A B. 3 A C. 1.75 A D. 1.33 A

Answers

First you calculate the total resistance. Since the resistors are connected in series, the total equivalent resistance is:
Rtotal = R1 + R2 = 10 + 4 = 14 ohms

Now you calculate the electrical current with Ohm's law:
I = U/Rtotal = 42/14 = 3A

Correct answer: B. 3 A

Which types of electromagnetic waves have higher frequencies than the waves that make up ultraviolet light? Check all that apply. -radio waves
-infrared light
-microwaves
-gamma rays
-visible light
-X-rays

Answers

On this list, 'gamma-'s and 'X-'s are the ones.

its gamma rays and x-rays they are the only two out of my knowledge.

Water pressure in a lake is greater __________. a. at the bottom b. at the surface c. at any point, for pressure is the same d. throughout midway to the bottom

Answers

Answer:

The correct answer  a

Explanation:

In the fluid pressure is a measure of force per unit area, in this case force is the weight of the fluid that increases as we descend. Therefore, the pressure increases with the depth of the fluid.

The correct answer  a

Answer:

a. at the bottom

Explanation:

Since water at the bottom of the lake is under pressure by all the water molecules above it plus atmospheric pressure. It is no doubt that the pressure at the bottom of the lake is larger than pressure in the middle and at the surface of the lake.

The velocity of a car traveling in a straight line increases from 0 meters/second to 30 meters/second in 8 seconds. What is the average acceleration of the car?

Answers

Answer : a=3.75\ m/s^2

Explanation :

Initial velocity of the car, u = 0 m/s

Final velocity of the car, v = 30 m/s          

Time taken, t = 8 s

Average acceleration of the car is defined as the change in velocity divided by time.

a=(v-u)/(t)

a=(30\ m/s-0)/(8\ s)

a=3.75\ m/s^2

Hence, this is the required solution.                                              

Average acceleration is how much the car increases (on average) per second.  So, since it increases by 30 m/s in 8 s, then dividing 30 by 8 will give you an average acceleration of 3.75 m/s^2 (remember that acceleration is ever increasing, so the unit is s^2, not just s)

Hope this helps!

A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?

Answers

IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy =  (mass)  x   (gravity) x (height)

                           = (0.15 kg) x (9.8 m/s²) x (20 m)

                           =      29.4 kg-m²/s²  =  29.4 joules .
The conservation of energy says that energy in an isolated system remains constant, that it is not created nor destroyed but transforms from one form of energy to another. 
So when the ball is thrown into the air, it has kinetic energy (KE). As it goes into the air, that kinetic energy transforms into gravitational potential energy (GPE). The higher it gets, the more GPE it has and therefore the less KE it has. 
At the top, all of it is transformed into GPE and there is no KE.
But total energy at any point after the ball is thrown until it lands it has the same energy throughout. 

Knowing this, you can set the beginning KE equal to the final GPE. 
KE = GPE

Insert formula for GPE.

KE = mgh
KE = 0.15kg*9.8m/s^2*20.0m
KE = 29 J

The ball initially had 29 J of kinetic energy