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Galina-37 [17]
3 years ago
5

In our lab experiment on Ohm's law, the power supply is connected to a circuit containing one resistor, and a direct current of

1.5 A is drawn. If you connect a second identical resistor to the power supply in parallel with the first, (similar to what we did in the lab) how much current is drawn from the battery?
(A) 3.0 A
(B) 0.75 A
(C) 1.5 A
(D) 6.0 A
Physics
1 answer:
lapo4ka [179]3 years ago
3 0

Answer: (A) 3.0=A

Explanation: In order to explain this problem we have to use the OHM law, given by: V=R*I

Besides, we have to consider the resitance equivalent for a parallel connection. This is given by:

1/Re=1/R1+1/R2

If we connect the same resistance, the equivalent resistance is R/2.

Initlally  the current is 1.5 A when one resistance is connected to the batttery. When a second resistance with the same value is connected in parallel to the battery, we have V=Re*Ifinal= (R/2)*Ifinal

also we know that V=R*Iinitial so Iinitial=V/R

then Ifinal= 2*V/R=2*Iinitial

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If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1,000 km
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The correct answer is:
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A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform
madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is \epsilon_{max}= 26.8 V

The emf induced at t = 1.00 s is \epsilon = 24.1V

The maximum rate of change of magnetic flux is   \frac{d \o}{dt}|_{max}  =26.8V

Explanation:

    From the question we are told that

        The number of turns is N = 44 turns

          The length of the coil is  l = 15.0 cm = \frac{15}{100} = 0.15m

          The width of the coil is  w = 8.50 cm =\frac{8.50}{100} =0.085 m

          The magnetic field is  B = 745 \ mT

          The angular speed is w = 64.0 rad/s

Generally the induced emf is mathematically represented as

        \epsilon = \epsilon_{max} sin (wt)

 Where \epsilon_{max} is the maximum induced emf and this is mathematically represented as

            \epsilon_{max} = N\ B\ A\ w

Where \o is the magnetic flux

            N is the number of turns

             A is the area of the coil which is mathematically evaluated as

             A = l *w

        Substituting values

           A = 0.15 * 0.085

               = 0.01275m^2

substituting values into the equation for  maximum induced emf

         \epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0

                 \epsilon_{max}= 26.8 V

 given that the time t = 1.0sec

substituting values into the equation for induced emf  \epsilon = \epsilon_{max} sin (wt)

      \epsilon = 26.8 sin (64 * 1)

        \epsilon = 24.1V

   The maximum induced emf can also be represented mathematically as

              \epsilon_{max} = \frac{d \o}{dt}|_{max}

  Where  \o is the magnetic flux and \frac{d \o}{dt}|_{max} is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

         \frac{d \o}{dt}|_{max}  =26.8V

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Answer:

Explanation:

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