What component in a circuit does this symbol represent?

You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated t

Nataly [62]

Answer:

Approximately $\rm 2.0\; V$ .

Approximately $\rm 30 \; mA$ . (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here $R_1= 10\; \Omega$ and $R_2= 100\; \Omega$ are connected in parallel. Their effective resistance would be equal to

$\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega$ .

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be $20\; \rm mA = 0.020\; \rm A$ . That should also be the current through the effective $\displaystyle \rm \frac{10}{11} \; \Omega$ resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

$V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V$ .

The voltage drop across the entire circuit would equal to

the voltage drop across the resistors, plus
the voltage drop across the LED.

In this case, that value would be equal to $1.83 + 0.182 \approx 2.0\; \rm V$ . That's the voltage that needs to be supplied to the circuit to achieve a current of $20\; \rm mA$ through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

$\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega$ .

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

$\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega$ .

Again, the current in a serial circuit is the same in all appliances.

$\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA$ .

7 0

3 years ago

100%

xxMikexx [17]

Answer:

1. The elastic potential energy is 0.0176 Joules

2. The kinetic energy of the pinball the instant it leaves the spring is 0.0176 Joules

3. The speed of the pinball the instant it leaves the spring is approximately 2.42212 m/s

4. The height of the part where the pinball is located on the machine above the ground is approximately 0.213 meters

Explanation:

The spring constant of the pinball machine's plunger, k = 22 N/m

The amount by which the pinball machine's plunger is compressed, x = 0.04 m

The mass of the pinball ball, m = 0.006 kg

1. The elastic potential energy, P.E. = 1/2·k·x²

By substitution, we get;

P.E. = 1/2 × 22 N/m × (0.04 m)² = 0.0176 J

The elastic potential energy, P.E. = 0.0176 J

2. At the instant the pinball leaves the spring, the plunger and therefore the force of the plunger no longer acts on the pinball

Since there are no external forces acting on the pinball to increase the speed of the pinball after it leaves the spring, the velocity reached is its maximum velocity, and therefore, the kinetic energy, K.E. is the maximum kinetic energy which by the conservation of energy, is equal to the initial potential energy

Therefore;

K.E. = P.E. = 0.0176 J

The kinetic energy of the pinball the instant it leaves the spring, K.E.= 0.0176 J

3. The kinetic energy, K.E., is given by the following formula;

K.E. = 1/2·m·v²

Where;

v = The speed or velocity of the object having kinetic energy K.E.

Therefore, from K.E. = 0.0176 J, and by plugging in the values of the variables, we have;

K.E. = 0.0176 J = 1/2 × 0.006 kg × v²

v² = 0.0176 J/(1/2 × 0.006 kg) = 88/15 m²/s²

v = √(88/15 m²/s²) ≈ (2·√330)/15 m/s ≈ 2.42212 m/s

The speed of the pinball the instant it leaves the spring, v ≈ 2.42212 m/s

4. The height of the pinball is given by the following kinematic equation of motion;

$v_h$ ² = u² - 2·g·h

Where;

$v_h$ = The velocity of the pinball at the given height = 1.3 m/s

u = v ≈ 2.42212 m/s (The initial velocity of the pinball as it the spring)

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height of the pinball above the ground

We get;

$v_h$ ² = 1.3² = 2.42212² - 2 × 9.8 × h

∴ h = (2.42212² - 1.3²)/(2 × 9.8) ≈ 0.213

The height of the part where the pinball is located on the machine above the ground, h ≈ 0.213 m

5 0

3 years ago

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Tcecarenko [31]

Answer:

(a) E=λ/(2\pi e0 r)

(b) E = 0

Explanation:

(a) We can use the Gaussian's Law to calculate the electric field at any distance r from the axis. By using a cylindrical Gaussian surface we have:

$\int \vec{E}\cdot d\vec{r}=\frac{\lambda}{\epsilon_o}$

where λ is the total charge per unit length inside the Gaussian surface. In this case we have that the Electric field vector is perpendicular to the r vector. Hence:

$E\int dr=E2\pi r=\frac{\lambda}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r \epsilon_o}$

(b) outside of the outer cylinder there is no net charge inside the Gaussian surface, because charge of the inner radius cancel out with the inner surface of the cylindrical conductor.

Hence, we have that E is zero.

hope this helps!!

3 0

3 years ago

Read 2 more answers

A wheel is spinning about its axle such that the wheel rotates by 2π radians (1 rev) every 0.43 s. While the wheel spins, the ax

GrogVix [38]

Answer:

τ = 166.67 N m

Explanation:

The torque is related to the angular momentum

τ = dL / dt

Bold indicates vectors, where tau is the torque and L is the angular momentum

In this case we put the variations, It's like looking for middle values

τ = ΔL / Δt

Let's calculate

τ = 50.00 / 0.30

τ = 166.67 N m

Applied in the direction of rotation at the end of the wheel

3 0

3 years ago

A leopard with a mass of 55.00 kg climbs 12.0 m up a tree. What is it’s gain in GPE

PSYCHO15rus [73]

The answer is: 6,468 j

4 0

3 years ago