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2 years ago

Find the value of currents through each branch

Physics

1 answer:

Irina-Kira [14]2 years ago

8 0

Answer:

the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

(I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

(I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

$\left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]$

These equations have the solution ...

$\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]$

This means the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

(I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

It helps to be familiar with the formulas for resistors in series and parallel.

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4 0

1 year ago

A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.

Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

$a=\dfrac{v-u}{t}$ .............(1)

Now, the second equation of motion is :

$s=ut+\dfrac{1}{2}at^2$

Put the value of a in above equation as :

$s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2$

$s=\dfrac{t(u+v)}{2}$

$u=\dfrac{2s}{t}-v$

$u=\dfrac{2\times 47}{8.6}-2.3$

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0

3 years ago

A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d

blsea [12.9K]

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster, m = 500.0 kg,

Displacement travelled by the dragster, s = 402 m,

Time taken in this travel, t = 5.0 s,

Final velocity of the dragster, v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation, s = ut + (1/2)at^2.

402 = u*5 + (1/2)*a*5^2

10*u + 25*a = 804. ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650. .............(2)

Solving (1)and (2), we get,

u = 30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W = 3987840. J

Therefore power rating of the dragster is given by,

P ⇒ W/t. = 3987840/5 = 797568 watt.

P ⇒ 797568/746 = 1069.1 hp.

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5 0

2 years ago

Read 2 more answers

How long will it take an object to move 100 meters if the object is traveling with and average speed of 0.5 meters per second

gogolik [260]

100\0.5 = 200 seconds

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2 years ago

while looking at a graph, you notice a period of time where the line is perfectly horizontal what is most likely taking place du

aev [14]

Answer:

where the y axis is

Explanation:

In more simple terms, a horizontal line on any chart is where the y-axis values are equal. If it has been drawn to show a series of highs in the data, a data point moving above the horizontal line would indicate a rise in the y-axis value over recent values in the data sample.

7 0

2 years ago