Monday  1 Feb:

 

Journal:  Two bulbs (5 ohms & 10 W) are first (Case 1) placed in a Parallel circuit and then (Case 2) in a Series circuit. (Case 3) A wire is placed as shown.

electricity III, diagram 5

 

 

 

Case 1:  (a) Which one is brighter?

(b) Which one uses more power?

 

Case 2:   (a) Which one is brighter?

 (b) Which one uses more power?

 

Case 3: What is the result?

 

 

 

 

 

Case 1:   (1) 5 W is brighter (more current/less resistance for same voltage?

 (2) 5 W uses more power (more current& same voltage) P = IV

 

 

Case 2:   (1) 10 W is brighter (uses more voltage because of higher resistance in order to send same current through line - both bulbs have same current going through them)

 (2) 10 W uses more power (same current, but uses higher voltage) P = IV

 

(3) Light go out (Short)

Parallel (Resistors):

 

 

 

 

 

 

 

 

 

Resistors (R1 & R2) 10 ohms (W) each.

 

 (Parallel) Requiv = (R1)(R2) / (R1) + (R2) 

            (may only use 2 resistors at a time with this formula)

 

 

 

 

 

 

In parallel circuits current is split at least one junction.3. In the diagram of a parallel circuit above, suppose R1 and R2 were light bulbs.

   a) If the light bulb at R2 burned out, what would happen to the brightness of the light bulb at R1? Why

  b) If the light bulb at R2 was replaced with a wire    (no resistor) , what would happen to the brightness of the light bulb at R1? Why

 

3.a) Nothing, Parallel

  b) R1 go out, Short - no resistance in wire.

 

 

 


4. Two resistors, 2.0 ohm and 3.0 ohm, are connected in parallel with a 12 V battery. Calculate: (a) the effective resistance of the circuit; (b) the current leaving the battery; (c) the current through each resistor

 

 

 

 

 

 

 

4.a)  (R1)(R2) /   (R1) + (R2) =Req1-2

 

(2)(3) /   (2) + (3)   =  Req1-2 = 1.2W

 

b) V = IR  --> 12 = I(1.2) --> I = 10A

c) V = IR  --> 12 = I(2) --> I = 6A

    V = IR  --> 12 = I(3) --> I = 4A

 

 

 

3. What is the equivalent resistance for the two parallel circuits above: a. 2 resistors (10 ohms (W) each)? b. 4 resistors (1 W, 6 W, 3 W,2 W))?

 

Four Resistors in Parallel

 

 

 

 

 

 

 

a.       Requiv = (R1)(R2) / (R1) + (R2)

                             Requiv = (10x10)/(10+10) = 5 W

 

b.       Requiv = (R1)(R2) / (R1) + (R2)

                             Requiv = (1x6)/(1+6) =  0.86W

          Requiv = (R1)(R2) / (R1) + (R2)

                             Requiv = (3x2)/(3+2) =  1.2W

Requiv = (R1)(R2) / (R1) + (R2)

                             Requiv = (1.2 x 0.86)/(1.2+0.86)

=  0.5W

 

4. If the total current is 24.0 amps, what is the voltage across the 4 resistor parallel circuit?

         

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ohms Law: V =IR  à V = (24)(0.5)

                                                            à V = 12.0 Volts

5. What is the current through each of the 4 resistors?

           

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 (1 W): V=IR à 12 =I(1)  à I =12.0 amps

        (6 W): V=IR à 12 =I(6)  à I =2.0 amps

(3 W): V=IR à 12 =I(3)  à I =4.0 amps

(2 W): V=IR à 12 =I(2)  à I =6.0 amps

                                      Total of 24.0 amps

 

 

 

 

 

 

 

 

 

 

 

Between equal 1Ohm's  resistors =  6V drop

After 2 Ohm's (Twice the resistance, twice the voltage drop) = 8V drop

Difference = 2 V (higher on top parallel line)

 

 

 

 

 

 

6. What is the current in each line

(I1, I2, & I3)?

 

 

 

 

 

 

       

  

 

                                                    B Loop                    A Loop

 

 

 

 

First solve as a voltage problem:

The Voltage around the two loops starting (at the dot) must = 0. Remember V = IR

a)  +40I1 + 4 = 0

+40I1 =  -4  --> I1=  - 0.1A

 

b)  + 12 + -80I2  -40I1 = 0

     + 12 + -80I2  -40(-0.1) = 0

                   -80I2 = -12 - 4  --> -16

                    I2 = 0.2A

c)  I2  = I1 + I3  

-->   0.2A = -0.1A + I3  

I3   =  0.3A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Cequiv = (C1 x C2)/ (C1)+(C2)

 

Capacitance (effective size of a capacitor):

C = Q/V (charge/volts)

How to increase capacitance (three things):

  1. Increase plate area (more charge can be stored).
  2. Decrease distance between plates (stronger force of attraction)
  3. Use better dielectric (holds more charge because doesn’t let charge go from plate to plate)

Co = EoAo/do

 

A

 

B

C

6. The voltage across the 5mF capacitor is?

 

 

100V (voltage across parallel lines are equal)

(in this case, it is equal to the Battery’s)

7. The voltage across the right side circuit (4mF, 2mF, & 3mF) is?

100V (voltage across parallel lines are equal)

(in this case, it is equal to the Battery’s)

8. The equivalent capacitance for the 4 microfarad (mF) and the 2mF capacitors is?

Parallel:

Cequiv = C1 + C2 à 4mF + 2mF = 6mF (6-microfarads)

9. What is the difference in potential between points:

a. A and B

b. B and C

c. C and A

a. Cequiv = C1 + C2 à 4mF + 2mF = 6mF

Parallel part’s (4mF + 2mF) has twice the capacitance as the 3mF capacitor. Therefore, it takes twice the voltage to keep its charge. Total voltage from A to C is 100V. So,

2/3 of 100V = 66.7V (from A to B)

b. Parallel part’s (4mF + 2mF) has twice the capacitance as the 3mF capacitor. Therefore, it takes twice the voltage to keep its charge. Total voltage from A to C is 100V. So,

1/3 of 100V = 33.3V (from B to C)

c. 100V (voltage across parallel lines are equal and voltage across a battery is the battery’s voltage)

10. The equivalent capacitance for the right side circuit (4mF, 2mF, & 3mF) capacitors is?

 

Series:

Cequiv = (C1 x C2)/ (C1)+(C2)

à (6 x 3) / (6+3) à 2mF

 

11. The equivalent capacitance for this network is most nearly

Parallel:

Cequiv = C1 + C2 à 2mF + 5mF = 7 mF

12. The charge stored in the 5-microfarad capacitor is most nearly

C = Q/V à Q = CV à Q = (5x10-6)(100)

Q = 5 x 10-4Coulombs

13. What is the PE of the 5-microfarad capacitor?

PEE = 1/2CV2 à (.5)(5x10-6)(100)2

PEE à .025J

Also PEE = 1/2QV à (.5)(5x10-4)(100) = .025J

  1. A parallel-plate capacitor has a capacitance C0. A second parallel-plate capacitor has plates with 10 times the area and 5 times the separation. The capacitance of the second capacitor is most nearly

    Co = EoAo/do

    C2 = Eo10Ao/5do

    C2 = 2Co

     

  2. Ten Joules of work is needed to move 2 coulombs of charge from one point to another with no change in velocity. Which of the following is true between the two points?

W = DVq

10 = DV(2) à the change in potential is

DV = 5 Volts or

W = DPE = q DV à (2)(5) =10 J

E = volts per meter (V/m)

à 5V/m is electric field strength

 

 

 

 

 

 

16. Two large parallel conducting plates -Q and Q are connected to a battery of emf 12V, as shown above. a. Does the force at various distances between the plates? b. Does the potential at various distances between the plates? If so, where is it the highest & lowest potential?

a. No. All the places between the plates have the same force & direction (uniform field).

b. Yes. The closer you are to the + plate, the higher the potential. It requires work to move a proton against the field (towards the + plate) or higher potential (giving it more PE). However, an electron will "fall" to the + plate (or higher potential) but it will lose PE. An electron must be moved with the field (toward the – plate/lower potential) to gain PE.