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Basile [38]
3 years ago
5

The rms current output in a circuit is 4.67 A. What is the maximum current?

Physics
2 answers:
jek_recluse [69]3 years ago
6 0

Answer:

I_O = \sqrt2 I_{rms}

I_O = \sqrt2 (4.67) = 6.6 A

Explanation:

As we know that maximum current in AC circuit is related to rms value of the current as per the equation below

I_{rms} = \frac{I_O}{\sqrt2}

here the value of maximum current or current amplitude is at numerator.

So in order to find maximum current or current amplitude we can say

I_O = \sqrt2 I_{rms}

now plug in the value of rms current in the above equation

I_O = \sqrt2(4.67)

I_O = 6.6 A

so maximum current is 6.6 A

Mila [183]3 years ago
5 0

Answer:Electrical power consumed by a resistance in an AC circuit is different to the power consumed by a reactance as reactances do not dissipate energy

   

In a DC circuit, the power consumed is simply the product of the DC voltage times the DC current, given in watts. However, for AC circuits with reactive components we have to calculate the consumed power differently.

Electrical power is the “rate” at which energy is being consumed in a circuit and as such all electrical and electronic components and devices have a limit to the amount of electrical power that they can safely handle. For example, a 1/4 watt resistor or a 20 watt amplifier.

Electrical power can be time-varying either as a DC quantity or as an AC quantity. The amount of power in a circuit at any instant of time is called the instantaneous power and is given by the well-known relationship of power equals volts times amps (P = V*I). So one watt (which is the rate of expending energy at one joule per second) will be equal to the volt-ampere product of one volt times one ampere.

Explanation:

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Please help me with this question ASAP.
victus00 [196]

Answer:

i. 6.923 V

ii. The e.m.f. = 22.5 V

Explanation:

i. The given parameters are;

Length of potentiometer = 1 m

The resistance of the potentiometer = 10 Ω

The e. m. f. of the attached cell = 9 V

The current, I flowing in the circuit = e. m. f/(Total resistance)

The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A

The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire

The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V

ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

\dfrac{E}{R_{balance}} = \dfrac{V}{R_{cell}}

Where:

E = e.m.f. of the balance point cell

R_{balance} = Resistance of 75 cm of potentiometer wire  = 0.75×10 = 7.5 Ω

R_{cell} = Resistance of the cell in the circuit = 3 Ω

V = e.m.f. attached cell = 9 V

\dfrac{E}{7.5} = \dfrac{9}{3}

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V

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