All categories

2 years ago

What's the voltage drop running through the parallel portion of the circuit?

Physics

2 answers:

Nady [450]2 years ago

5 0

The answer is 18V ( I attached my working for you):
Basically I found the resistance of the parallel and the series to give the equivalent resistance, from there I calculated the current.
I used that to calculate the voltage drop over the parallel circuit.

Leni [432]2 years ago

4 0

R 3/4 = (R3 * R4) / (R3 + R 4) = ( 9 * 18 ) /(9 + 18 ) = 162 / 27 = 6 Ohms
R e = R 1 + R 2 + R 3/4 + R 5 = 3 + 6 + 6 + 15 = 30 Ohms
I = U / Re = 90 V / 30 Ohms = 3 A
Finally for the voltage U 3/4 ( the parallel portion of the circuit ):
U 3/4 = 6 Ohms * 3 A = 18 V
Answer: 18 V

You might be interested in

What is the difference between charles law and boyle's law?

Leokris [45]

Boyle's law talks about the relationship between pressure and volume (high pressure = low volume, and vice-versa), while Charles's law talks about the relationship between volume and temperature (high temperature = high volume, and vice-versa).

3 0

2 years ago

g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal r

Naddika [18.5K]

Answer:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

6 0

2 years ago

A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along

Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

$F(x)=(3\ N/m^2)x^2+(6\ N/m)x$

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

$W=\int\limits {F{\cdot} dx}$

$W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J$

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

3 0

2 years ago

Read 2 more answers

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

aleksandrvk [35]

Hope this helps!

-Lilly

3 0

2 years ago

PLEASE ANSWER WILL GIVE BRAINLIEST

Karolina [17]

Answer:

B. NET force: 2 resultant motion: left

ItsOniiSama avatar

C. Net force: 3 Resultant motion: Left

ItsOniiSama avatar

D. Net Force: 7 Resultant motion: right

ItsOniiSama avatar

E. Net Force:0 resultant motion: NO MOTION

ItsOniiSama avatar

F. NET Force: 3 resultant motion: Down

ItsOniiSama avatar

G. NET FORCE: 10 resultant motion: up

ItsOniiSama avatar

H. Net force: 3 Resultant motion: left

ItsOniiSama avatar

I. Net force: 50 Resultant motion: right

ItsOniiSama avatar

J. NET FORCE: 75 Resultant motion: down

ItsOniiSama avatar

K. Net force :200 Resultant motion: Right

ItsOniiSama avatar

L. Net force: 0 resultant motion:No motion

Explanation:

8 0

2 years ago