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Lesechka [4]
2 years ago
14

Consider a circuit with a main wire that branches into two other wires. If the current is 10 A in the main wire and 4 A in one o

f the branches, how much current is in the other branch
Physics
1 answer:
k0ka [10]2 years ago
4 0

If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.

According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.

Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.

Consequently, the second wire's power may be expressed as;

I = I1+ I2 [ where I= total current (10A);

                I1= current in one branch (4A) &

                I2= current in another branch]

⇒I2 = I - I1

⇒I2 = 10A - 4A

⇒I2 = 6A

Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.

Learn more about Kirchhoff's law here:

brainly.com/question/6417513

#SPJ4

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A generator produces 38 mwmw of power and sends it to town at an rms voltage of 78 kvkv. part a what is the rms current in the t
4vir4ik [10]

The rms current in the transmission lines is I = 487.18 A.

The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force  is used to represent the source. it is the rectangular root of the time average of the voltage squared.

Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.

Electric power is  by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values

power = 38 M watt

rms voltage = 78 K v

power = IV

I = power/V

I = (38 * 1000000)/78*1000

I = 487.18 A.

Learn more about rms current here:-brainly.com/question/20913680

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2 years ago
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yaroslaw [1]

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Answer:

v = 876 m/s

Explanation:

It is given that,

Number of mol of Neon is 2 mol

Temperature, T = 308 K

Mass, m = 0.02 kg

Value of R - 8.31 J/mol-K

We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,

\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s

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