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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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Answer:

Diameter of the piston would be 0.71 m (71.1 cm)

Explanation:

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\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

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So that:

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A_{1} = \frac{2903.57*0.00334}{24.41}

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A_{1} = \pi r^{2}

0.3973 = \frac{22}{7} x r^{2}

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r = \sqrt{0.1264}

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r = 0.36 m

Diameter of the piston = 2 x r

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Answer:

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