a) A circuit consists of a 12 V battery connected to a switch, which is connecte

Question

a) A circuit consists of a 12 V battery connected to a switch, which is connected to a light bulb of resistance 120 , and then back to the battery. If the switch is open, which statement is true of I (the current through the switch) and Vs (the difference in potential across the poles of the switch)?

I = 10 A, Vs = 0 V I = 0 A, Vs = 0 V     I = 0 A, Vs is more than 12 V I = 10 A, Vs is not zero but less than 12 V I = 0.1 A, Vs = 12 V I = 10 A, Vs is more than 12 V I = 0.1 A, Vs is more than 12 V I = 0 A, Vs = 12 V I = 0.1 A, Vs = 0 V I = 10 A, Vs = 12 V I = 0 A, Vs is not zero but less than 12 V I = 0.1 A, Vs is not zero but less than 12 V

b) A long wire (with measurable resistance) is connected to a battery of potential difference V, and current I flows through the wire. The voltage is now doubled, but a fan keeps the the temperature of the wire from increasing. What will happen to the current I and the resistivity of the wire ?

I will remain the same, will quadruple I will double, will quadruple     I will halve, will be quartered I will double, will be quartered I will halve, will remain the same I will remain the same, will double I will halve, will halve I will halve, will double I will double, will double I will remain the same, will be quartered I will remain the same, will remain the same I will double, will remain the same I will halve, will quadruple I will double, will halve I will remain the same, will halve

c) Seven identical wires of measurable resistance are attached consecutively: one end of the first wire is connected to the one end of the second wire; the other end of the second wire is connected to one end of the third wire, and so on. The free end of the first wire is connected to the positive terminal of a battery of potential difference V, and the free end of the seventh wire is connected to the negative terminal of the battery. The potential difference across wire 1 is V1, and the current flowing through wire 1 is I1, If an eighth identical wire is introduced between the seventh wire and the battery, what happens to V1 and I1?

V1 decreases, I1 remains the same Both V1 and I1 increase     V1 remains the same, I1 increases V1 increases, I1 remains the same V1 remains the same, I1 decreases V1 decreases, I1 increases V1 increases, I1 increases Both V1 and I1 remain the same Both V1 and I1 decrease

d) Now, seven identical wires of measurable resistance are connected across the the terminals of the battery of potential difference V. (In other words: one end of each wire is connected to the positive terminal of the battery, and the other end is connected to the negative terminal of the battery.) The potential difference across wire 1 is V1, and the current flowing through wire 1 is I1, If an eighth identical wire is now added so it connected like the others (across the battery), what happens to V1 and I1?

V1 decreases, I1 remains the same V1 increases, I1 remains the same     Both V1 and I1 decrease Both V1 and I1 remain the same Both V1 and I1 increase V1 remains the same, I1 decreases V1 increases, I1 increases V1 decreases, I1 increases V1 remains the same, I1 increases

Watch out for units!

Suppose a 53.9 V battery is connected to a long cylindrical wire, and a current of 4.74 A flows.

a) Find the number of electrons that pass any given point in the wire every second.
electrons

b) Find the total resistance of the wire.
RTOT =

c) If the wire is 30 m long, and its diameter is 0.0741 cm, find the resistivity of the wire.
= -m

d) Find the work done by this battery to push one electron through the entire length of the wire.
W = J

e) You will have three chances for this question.
A cylindrical wire of length L has resistance R = 888 . The wire is now passed through a die so it is drawn out to length 2.92L. What is the new resistance of the wire?
HINT: The length is not the only quantity that has changed!
R =

Explanation / Answer

a)

Current = Charge/time

So, Charge/time = 4.74 A

Now, total charge = n*e

where e = charge of each electron = 1.6*10^-19 C

So, 4.74 = n*1.6*10^-19

So, n = 2.96*10^19 electrons <-----answer

b)

Rtot = V/I = 53.9/4.74 = 11.37 ohm

c)

= R*A/L = 11.37*pi*(0.0741*10^-2/2)^2/30 = 1.63*10^-7 ohm.m

d)

Work done by battery,

W = V*I = 53.9*4.74 = 255.5 J

e)

Let the radius of cross section of the wire after passing through die be r

Now, final and initial volumes must be same

So, (4/3)*pi*(R)^2*L = (4/3)*pi*r^2*2.92*L

So, r = 0.585*R

Inital Resistance, R =  *L/(pi*R^2) = 888 ohm

New resistance, R' = *(2.92*L)/(pi*r^2) = 2.92/(0.585)^2*R

So, R' = 8.53*R = 8.53*888 = 7571.4 ohm <------answer

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