MULTIPLE CHOICE, Choose the one alternative that best completes the statement or

Question

MULTIPLE CHOICE, Choose the one alternative that best completes the statement or answers the question. 1) A 50- ? resistor and a 9.0-? resistor are connected in parallel. A 40-? resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through the 5.0-2 resistor? A) 0.30A B) 1A C) 0.53A D) 0.83 A 2) An uncharged 1.0-uF capacitor is connected in series with a 234? resistor, an ideal 70-V battery, and an open switch. What is the voltage across the capacitor 11 ms after closing the switch? A) 2.7 V B) 2.6 V C) 1.6 V D) 0.62 V 3) A cylindrical wire has a resistance R and resistivity p. If its length and diameter are BOTH cut in half, what will be its resistance? A) 4R B) R2 C) R/4 D) 2R 4) An electric device delivers a current of 5.0 A to a device. How many clectrons flow through this device in 10 s? (e-1.60 x 10-19 C) B) 31 × ?0zu A)3.1 × C) 0.20 D) 2.0 E) 20 104 5) The charge on the square plates of a parallel-plate capacitor is 0. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to A) 20 C) 02. D) Q14 E) 40 6) A charged capacitor stores energy U. Without connecting this capacitor to anything, dielectric having dielectric constant K is now inserted between the plates of the capacitor, completely filling the space between them. How much energy does the capacitor now store? A) U B) KU C) D)2KU E) 2K

Explanation / Answer

1)

The potential difference is constant in parallel combination.

The equivalent resistance in the circuit is

Req = {(1/(5) + 1/(9)}^-1 + 4

= 7.214 ohm

So the current in the circuit is

I = V/Req

I = (6 V)/(7.214 ohm)

I = 0.83 A

The voltage across parallel combination is

V' = (0.83 A )(3.214 ohm)

= 2.673 V

Therefore, the current in the 5 ohm resistor is

I' =(2.673 V) / (5 ohm)

I' = 0.53 A

2)

? = RC = 23 x 10^3 ohm x 1 x 10^-6 F

= 0.023 s

V = Vo(1 - e^(-t/?) )

V = 7(1 - e^(-0.011/0.023) )

V = 2.661 V = 2.7 V (approx)

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