Math, 15.05.2020meteor13

Four ideal 12 batteries (no internal resistance) are connected in the following configuration;
Two group with each group have 2 batteries connected in series. The resulting two groups are then connected in parallel. If a 2000 ohms load is connected across the resulting circuit, determine the following:
1. Draw the circuit diagram
2. What voltage seen in the load?
3. What is the current seen by the load?
4. What power is dissipated by the lead?

Answers

Explanation:

An ever-present challenge in electronic circuit design is selecting suitable components that not only perform their intended task but also will survive under foreseeable operating conditions. A big part of that process is making sure that your components will stay within their safe operating limits in terms of current, voltage, and power. Of those three, the “power” portion is often the most difficult (for both newcomers and experts) because the safe operating area can depend so strongly on the particulars of the situation.

Explanation:

a. Since the resistors are connected in series,

Req = 10 + 14 = 24 Ω

b. By Ohm's Law

I = V / Req

I = 120 / 24

I = 5 A

c. P = I^2 × Req

P = 5^2 × 24

P = 600 W

Electric power (P) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅V= 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power P = IV = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (1 kA ⋅V = 1 kW). To see the relationship of power to resistance, we combine Ohm’s law with P = IV. Substituting I = V/R gives P = (V/R)V=V2/R. Similarly, substituting V = IR gives P = I(IR) = I2R. Three expressions for electric power are listed together here for convenience:

answer:

B.pohh

Explanation:

yun pohh

#carry onlerner

answer:

1.one path

A series circuit is wired with only one path for the current to flow through all the devices in a row and back to the starting point. The same current flows through each part of a series circuit. If the circuit is broken at any point there won't be any current that will flow.

2.Increasing the voltage increases the brightness of the bulb. When a bulb in a series circuit is unscrewed all bulbs in the circuit go out. Increasing the number of bulbs in a series circuit decreases the brightness of the bulbs. In a series circuit, the voltage is equally distributed among all of the bulbs.

3.The same current flows through each part of a series circuit. The total resistance of a series circuit is equal to the sum of individual resistances. ... The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor. If the circuit is broken at any point, no current will flow.

4.Increasing the number of bulbs in a series circuit decreases the brightness of the bulbs. In a series circuit, the voltage is equally distributed among all of the bulbs. Bulbs in parallel are brighter than bulbs in series. In a parallel circuit the voltage for each bulb is the same as the voltage in the circuit.

5.In a series circuit, the sum of the voltages consumed by each individual resistance is equal to the source voltage. ... In a series circuit, the current that flows through each of the components is the same, and the voltage across the circuit is the sum of the individual voltage drops across each component.

6.If a light switch is on to a busted bulb, does it still use electricity? For a standard incandescent bulb, the answer is no with one exception. If the glass is broken but the filament remains intact, the bulb will use electricity and the filament, though short lived, will glow until it burns up.

7.To find out, we need to be able to calculate the amount of power that the resistor will dissipate. If a current I flows through through a given element in your circuit, losing voltage V in the process, then the power dissipated by that circuit element is the product of that current and voltage: P = I × V

8.The three types of thermal energy transfer are conduction, convection and radiation. Conduction involves direct contact of atoms, convection involves the movement of warm particles and radiation involves the movement of electromagnetic waves.

9.A circuit is the path that an electric current travels on, and a simple circuit contains three components necessary to have a functioning electric circuit, namely, a source of voltage, a conductive path, and a resistor. Circuits are driven by flows.

CORRECT ME IF I'M WRONG:)

answer: D

Explanation:0.960 volts

And so the voltage in the he secondary will be that of the primary times the secondary turns divided by primary turns. So it's 120 volts times four divided by 500 which gives 0.960 volts.

R=5.92\pm0.02\:\Omega=5.94\:\Omega\:\textrm{or }5.9\:\Omega\\I=1.40\pm0.05\:\textrm{mA}=1.45\:\textrm{mA or }1.35\:\textrm{mA}\\

Then,

P=I^2R=(1.45\:\textrm{mA})^2(5.94\:\Omega)=12.49\:\mu\textrm{W}\\P=I^2R=(1.35\:\textrm{mA})^2(5.9\:\Omega)=10.75\:\mu\textrm{W}

To write it into uncertainty, first we take the average of the power:

\dfrac{12.49+10.75}{2} =11.62

and then the uncertainty is

12.49-11.62=0.87

Therefore,

P=11.62\pm0.87 \:\mu\textrm{W}



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