qué le sucede a la intensidad de la luz de cada bombillo en un circuito en serie si agregas más bombillos al circuito?

The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A

Bogdan [553]

Answer:

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4(7.4+R)... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4(7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R= 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms

5 0

3 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

3 years ago

Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at

beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0

3 years ago

Which biome has the most extreme conditions

Lelu [443]

Well this all depends on the region you would like to know about. One biome would be The Tundra. This biome is a very bitter cold. Some times the temperature can drop to -45f! So your answer more than likely would be Tundra.

Have a wonderful day user!

3 0

3 years ago

Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third

seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒ F₄ = - (F₁ + F₂ + F₃)

⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0

3 years ago