Answer:
* energy is proportional to masses. in a graph it would look like a line
* kinetic energy varies with the square of the velocity, In a graph it gives rise to a quadratic curve
Explanation:
Kinetic energy is defined by
K = ½ m v²
when analyzing this expression we can see:
* energy is proportional to masses. Therefore, doubling the mass doubles the kinetic energy and if the mass rises 4 times the energy rises 4 times, that is, they are directly proportional, in a graph it would look like a line
* kinetic energy varies with the square of the velocity. Therefore by doubling the speed the energy goes up 4 times. In a graph it gives rise to a quadratic curve
The direction of current flow is given by an arrow in the electrical circuit.
Several categories are used to categorise electrical circuits. One-way current only flows through a direct-current circuit. As with most residential circuits, an alternating-current circuit is a channel for transmitting electrical current that pulses back and forth numerous times each second.
A battery or generator, a device that provides energy to the charged particles that make up the current, a device that uses current, such as a lamp, an electric motor, or a computer, and the connecting wires or transmission lines make up an electrical circuit.
A source of electrical power, two wires that can transmit current, and a light bulb make up an electrical circuit. Both wires have one end connected to a cell terminal and the other end connected to a light bulb. When the light bulb is turned off, the electrical circuit is disrupted.
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Answer:
(a). The rms load current is 1.08 A.
(b). The rms primary current is 0.108 A.
Explanation:
Given that,
Primary voltage = 120
Secondary voltage = 12
Load = 10 ohm
We need to calculate the number of turns in primary and secondary coil
Using formula of voltage
![\dfrac{V_{1}}{V_{2}}=\dfrac{N_{1}}{N_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7B1%7D%7D%7BV_%7B2%7D%7D%3D%5Cdfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D)
Put the value into the formula
![\dfrac{N_{1}}{N_{2}}=\dfrac{120}{12}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%3D%5Cdfrac%7B120%7D%7B12%7D)
![\dfrac{N_{1}}{N_{2}}=10](https://tex.z-dn.net/?f=%5Cdfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%3D10)
![\dfrac{N_{2}}{N_{1}}=0.1](https://tex.z-dn.net/?f=%5Cdfrac%7BN_%7B2%7D%7D%7BN_%7B1%7D%7D%3D0.1)
We need to calculate the resistance
Using formula of resistance
![R=10\times(\dfrac{N_{1}}{N_{2}})^2](https://tex.z-dn.net/?f=R%3D10%5Ctimes%28%5Cdfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%29%5E2)
Put the value into the formula
![R=10\times(10)^2](https://tex.z-dn.net/?f=R%3D10%5Ctimes%2810%29%5E2)
![R=1000\ \Omega](https://tex.z-dn.net/?f=R%3D1000%5C%20%5COmega)
We need to calculate the current in primary coil
Using formula of current
![I_{1}=\dfrac{120-12}{1000}](https://tex.z-dn.net/?f=I_%7B1%7D%3D%5Cdfrac%7B120-12%7D%7B1000%7D)
![I_{1}=0.108\ A](https://tex.z-dn.net/?f=I_%7B1%7D%3D0.108%5C%20A)
We need to calculate the current in secondary coil
Using formula of current for secondary coil
![\dfrac{I_{1}}{I_{2}}=\dfrac{N_{2}}{N_{1}}](https://tex.z-dn.net/?f=%5Cdfrac%7BI_%7B1%7D%7D%7BI_%7B2%7D%7D%3D%5Cdfrac%7BN_%7B2%7D%7D%7BN_%7B1%7D%7D)
![I_{2}=I_{1}\times\dfrac{N_{1}}{N_{2}}](https://tex.z-dn.net/?f=I_%7B2%7D%3DI_%7B1%7D%5Ctimes%5Cdfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D)
Put the value into the formula
![I_{2}=0.108\times10](https://tex.z-dn.net/?f=I_%7B2%7D%3D0.108%5Ctimes10)
![I_{2}=1.08\ A](https://tex.z-dn.net/?f=I_%7B2%7D%3D1.08%5C%20A)
Hence, (a). The rms load current is 1.08 A.
(b). The rms primary current is 0.108 A.
The value of the current in each arm of the circuit is given by Kirchhoff's
Rules.
- The correct option for (I₁, I₂) is; b-.
![\underline{(1.33, \ 0.17)}](https://tex.z-dn.net/?f=%5Cunderline%7B%281.33%2C%20%5C%200.17%29%7D)
Reasons:
By Kirchhoff's junction rule, we have that the sum of current at a junction is given as follows;
Which by the direction of the currents in the given circuit diagram, we have;
Therefore;
I₁ = I₃ + I₂
According to Kirchhoff's loop rule theory, we have;
In the loop having the 2 Volts emf., we have;
-I₃·R₁ + 2 + I₂·R₂ = 0
I₃ = 1.17 A
R₁ = 2 Ω
R₂ = 2 Ω
Which gives;
-1.17 × 2 + 2 + I₂×2 = 0
I₂ × 2 = 2.34 - 2 = 0.34
I₂ = 0.34 ÷ 2 = 0.17
I₂ = 0.17 A
From the loop having the 7 Volts emf. we have;
7 - 0.17×2 - I₁ × 4 - I₁ × 1 = 0
Which gives;
7 - 0.34 - 5·I₁ = 0
5·I₁ = 7 - 0.34 = 6.66
I₁ = 6.66 ÷ 5 = 1.332
Therefore;
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