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Eduardwww
2 months ago
6

What is the differential equation governing the growth of current in the circuit as a function of time after t=0? express the ri

ght-hand side of the differential equation for di(t)dt in terms of i(t), vb, r, and l?
Physics
1 answer:
Yuliya22 [3.3K]2 months ago
3 0

Answer:

v_{b}=ir+L\frac{di}{dt}

Explanation:

A differential equation containing a term with di(t)/dt appears in an RL circuit. Here we express

v_{b}=v_{r}+v_{i}

where vr represents the voltage across the resistance, vi indicates the voltage across the inductance, and vb signifies the source voltage. Additionally, we have that

v_{r}=ir\\v_{i}=L\frac{di}{dt}

where L refers to the circuit's inductance, r is the resistance, and i is the current. By substitution, we obtain the differential equation

v_{b}=ir+L\frac{di}{dt}

I hope this is of benefit to you

regards

You might be interested in
Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.
serg [3582]

Answer:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

Consequently, we find that:

V_A = \frac{1}{4} V_B

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

L_A = L_B =L both wires share the same length

both wires carry an identical currentI_A = I_B =I

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

n_A = n_B =n

We also know that r_A = 2 r_B where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B

A_B = \pi r^2_B

Thus, we conclude that

A_A = 4 A_B

Now we are aware that the drift velocity of an electron in a wire can be described by:

v_d = \frac{I}{neA}

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

So we observe that:

V_A = \frac{1}{4} V_B

Thus, the most fitting answer is:

a. vA = vB/4

6 0
2 months ago
A particular brand of gasoline has a density of 0.737 g/mLg/mL at 25 ∘C∘C. How many grams of this gasoline would fill a 14.9 gal
Yuliya22 [3333]

Answer:

The answer to your inquiry is Mass = 41230.7 g or 41.23 kg.

Explanation:

Data

Density = 0.737 g/ml

Mass = ?

Volume = 14.9 gal

1 gal = 3.78 l

Process

1.- Convert gallons to liters

1 gal ---------------- 3.78 l

14.9 gal ------------- x

x = 56.44 l

2.- Convert liters to milliliters

1 l ------------------- 1000 ml

56.44 l --------------- x

x = (56.44 x 1000) / 1

x = 56444 ml

3.- Calculate the mass

Formula

Density = \frac{mass}{volume}

Solving for mass

Mass = density x volume

Substituting values

Mass = 0.737 x 56444

Result

Mass = 41230.7 g or 41.23 kg.

3 0
4 months ago
A suspension system is being designed for a 1300 kg vehicle. When the vehicle is empty, its static deflection is measured as 2.5
Keith_Richards [3271]

Answer: damping coefficient = 1.5×10^5Ns/m

Explanation:

Refer to the attached file for the solution

3 0
3 months ago
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