Engineering High School

Find the current and voltage through R₂ and R3

10V
R₁
30
R₂
50
R3
20

R4
50

Answers

Answer 1

The current through R2 is 0.3 A and the voltage across R2 is 15 V. The current through R3 is 0.1 A, and the voltage across R3 is 5 V.

What are current and voltage?

Generally, To find the current through R2 and R3 and the voltage across them, you can use Ohm's law and Kirchhoff's laws.

Ohm's law states that the current through a resistor is directly proportional to the voltage across it, with the proportionality constant being the resistance. Mathematically, this can be written as:

I = V / R

Where I is the current, V is the voltage, and R is the resistance.

Kirchhoff's laws can also be used to solve for the currents and voltages in a circuit. Kirchhoff's current law states that the sum of all the currents entering a node (a point where three or more branches meet) must be equal to the sum of all the currents leaving the node. Kirchhoff's voltage law states that the sum of the voltage drops around a loop must be zero.

Using these laws, you can set up a system of equations to solve for the unknowns in the circuit.

For the given circuit, you can start by labeling the currents and voltages as shown:

I1 = current through R1 I2 = current through R2 I3 = current through R3

V1 = voltage across R1 V2 = voltage across R2 V3 = voltage across R3

Then, you can use Ohm's law to write equations for the voltages across each resistor in terms of the currents and resistances:

V1 = I1 * R1 V2 = I2 * R2 V3 = I3 * R3

You can also use Kirchhoff's current law to write an equation for the current at the top node:

I1 + I2 + I3 = 0

And Kirchhoff's voltage law to write an equation for the voltage around the loop:

V1 + V2 + V3 = 0

Substituting the expressions for V1, V2, and V3 from Ohm's law into the equations from Kirchhoff's laws, you get a system of three equations with three unknowns:

I1 * R1 + I2 * R2 + I3 * R3 = 0 I1 + I2 + I3 = 0

You can then solve this system of equations to find the values of I1, I2, and I3, and then use Ohm's law to find the voltages V1, V2, and V3.

Solving this system of equations, you find that:

I1 = -0.2 A I2 = 0.3 A I3 = 0.1 A

V1 = -6 V V2 = 15 V V3 = 5 V

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Related Questions

Which of the foll?
Select the best option.

Anchor


Guardrails


Personal fall arrest systems


Safety nets

Answers

If you can't completely eliminate the risk, the safest and most reliable kinds of protection include guardrails, skylight screens, and other passive fall protection measures.

What is the best fall protection system?Although fall arrest systems are more suitable in some applications when personnel are compelled to operate near the leading edge, fall restraint systems are still preferred.If you can't completely eliminate the risk, the safest and most reliable kinds of protection include guardrails, skylight screens, and other passive fall protection measures.We advise operating in restraint if you are unable to prevent a hazard passively (with a railing, skylight screens, etc.).An employee who falls from a working level is stopped by a system known as a "personal fall arrest system."It is made up of an anchorage, connectors, a body belt or harness, a lanyard, a deceleration device, a lifeline, or suitable combinations of these, as well as connectors.

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As a technician, a simple test to check for excessive engine wear is to take a white paper towel and wipe
the oil from the
on it.

Answers

Crude oil comes in a variety of hues, including filthy yellow, dark brown, and black. It might be a thick substance that resembles oil or tar, or be thin like petrol. Ground crude oil is extracted and processed into a variety of goods, including fuel for use in diesel and petrol vehicles.

What to check for excessive engine wear?

Where an engine experiences the most wear is during starts and short rides. When an engine is cold, the surfaces are under-lubricated, and it takes a few seconds for the oil to fully cover them.

Therefore, As a technician, a quick test to look for excessive engine wear is to rub some oil from a white paper towel onto it. Fuel for use in diesel and petrol vehicles is one of the many products made from ground crude oil after it has been extracted and processed.

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Double filament bulbs can be used as the lone bulb in the stoplight circuit, taillight circuit, and the turn signal circuit.
True

False

Answers

True , Double filament bulbs can be utilized as the only bulb in stoplight, taillight, and turn signal circuits, which proves the given statement.

What is meant by double filament bulb?A halogen bulb having two light filaments arranged side by side is what is known as a dual filament bulb. Your current bulb probably has two filaments if your car utilizes a single bulb for both high and low beams.To convert this type of halogen bulb to a HID, you must first understand dual filament halogen lights (or LED).Dual filament bulbs like the 3157 are included in dual purpose lights. Each filament can be included into a separate circuit.The circuits for brake and tail lights in vehicles are where dual purpose bulbs are most frequently used. In automobiles like the PT Cruiser, the same bulb socket serves as both the brake and taillights.

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11. Which of the following is a type of file commonly used in the automotive industry?

A) Square file
B)Warding file
C) Both A and B
D)None of the above

Answers

The option that is type of file commonly used in the automotive industry is "Option C) which is Square files and Warding Files.

What is the definition of the above files?

A square file is a hand tool with a rectangular cross-section and flat edges, used for smoothing or shaping metal or wood. It is often used in the automotive industry for filing and shaping metal parts.

A warding file is also a hand tool, similar in shape to a square file but with a narrow, triangular cross-section.

Note that both square files and warding files are valuable tools for a variety of tasks in the automotive industry, including shaping and smoothing metal parts, filing down rough edges, and shaping keyways and other narrow slots.

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A steel elevator cable supports a load of 900 kg. The cable has a diameter of 2 cm and an initial length of 24 m. Find the stress and the strain of the cable and the amount it stretches under the load.

Answers

The stress and the strain of the cable and the amount it stretches under the load.

What is stress?

Stress is a natural response to life events that causes physical, emotional, and mental strain. It is the body’s way of responding to any kind of demand or threat. When a person experiences stress, their body releases hormones such as adrenaline and cortisol in order to help them respond to the challenge. This can cause a range of physical and mental effects, including increased heart rate, increased alertness, and a heightened state of awareness. Stress can be both positive and negative. Positive stress can help people stay focused and motivated, while negative stress can lead to feeling overwhelmed and anxious. It is important to find ways to manage stress in order to maintain physical and mental health. This can include healthy lifestyle habits such as regular exercise, social activities, and relaxation techniques like yoga and meditation. Additionally, talking to friends, family, or a mental health professional can help to reduce the impact of stress.

The stress, σ, in a cable is given by the equation:

σ = F/A

where F is the force the cable is supporting and A is the cross-sectional area of the cable. In this case, the force is 900 kg, or 9,000 N, and the cross-sectional area is πr2, where r is the radius of the cable. The radius of a 2 cm diameter cable is 1 cm, so the cross-sectional area is π(0.01)2 = 0.000314 m2.

Therefore, the stress in the cable is 9,000 N/0.000314 m2 = 28,621 N/m2.

The strain, ε, in a cable is given by the equation:

ε = ΔL/L

Where ΔL is the change in the length of the cable and L is the original length. In this case, the original length is 24 m and the amount the cable stretches under the load is ΔL = 900 kg × 9.81 m/s2 × 24 m = 21,216 kg m/s2.

Therefore, the strain in the cable is 21,216 kg m/s2/24 m = 884 kg m/s2.

Finally, the amount the cable stretches under the load is ΔL = 900 kg × 9.81 m/s2 × 24 m = 21,216 kg m/s2.

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