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Paladinen [302]
3 years ago
14

What help in reversing direction of current of current​

Physics
1 answer:
Novay_Z [31]3 years ago
3 0

Answer:

<em>To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection.  </em>

Explanation:

For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.

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A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

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Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

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V_{L} = - IwLsin(wt)

w = 500 rad/s

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Current:

I = 11.0 V / wL

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 = 11.0 / 250

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Now

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    = (0.044 A) (93 Ω)

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Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V_{R} = 2.05 V

8 0
3 years ago
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