Friday 18 Feb:

 

Concept(s) Today:  3rd motion equation (start L#47 & 48)

 

L#44: 1, 2, 5-14, 16, 20 (due Today 18 Feb)  (Waves)

 

Mon (28 Feb): EC for 5 min Physics Clip (Must have paper with Net address)

 

 

L#47: 1- 4, 7, 9, 11, 13 - 16 (due Tues 1 Mar)  (Kirchhoff's Laws)

 

L#48: 1- 4, 7-9, 11-14, 18, 19 (due Wed 3 Mar)  (3rd Motion Equation)

 

Lab Thurs 4 Mar

 

L#49: 1- 4, 7, 11-16, 18, 20 (due Fri 4 Mar)  (Circuit Analysis)

 

 

 

Review Monday 7 Mar

Test: 42-49 (Tues 8 Mar)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3rd Motion Equation:  (No time in this one)

 

2aDx = Vf2-Vi2

 

 

 

Kirchhoff's Law:

The current entering any junction is equal to the current leaving that junction. i1 + i4 = i2 + i3
 

The current entering any junction is equal to the current leaving that junction. i1 + i4 = i2 + i3

The sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

 

 

Example:

A circuit consists of a cell of emf 12.0 V in Parallel with a R1 = 2.0 W  connected in series with R2 = 3.0 W, and in parallel with R3 =  6.0 W.  

Determine:

  1. The voltage drop across the 6.0 W resistor

  2. The current through the 6.0 W resistor

  3. The voltage drop across the 2.0 W  and the 3.0 W resistor (in total).

  4. The resistance across the 2.0 W  and the 3.0 W resistors (in total).

  5. The current through the 2.0 W resistor

  6. The current through the 3.0 W resistor

  7. The voltage drop across the 2.0 W resistor

  8. The voltage drop across the 3.0 W resistor

  9. The current through the battery (I)

 

 

Solution

  1. The voltage drop across the 6.0 W resistor = Battery Voltage -->  V3 = 12 Volts

  2. The current through the 6.0 W resistor   V = I3R3:    12 = (I1)6 --> I3 = 2 amps

  3. The voltage drop across the 2.0 W  and the 3.0 W resistor (in total).  = Battery Volts: V1/2 = 12V

  4. The resistance across the 2.0 W  and the 3.0 W resistors (in total). series resistance: add -->R1/2 = 5.0 W

  5. The current through the 2.0 W resistor.  Current that goes through I1 is equal to current that goes through I2:            V1/2 =  I1R1/2 -->  12 = (I1/2)5 -->  I1/2 = I1= 2.4 amps

  6. The current through the 3.0 W resistor. V1/2 = IR1/2:    12 = (I2)5 --> I1/2 = I2= 2.4 amps

  7. The voltage drop across the 2.0 W resistor. V1 = I1R1:    V1 = (2.4)2 --> V1 = 4.8 volts

  8. The voltage drop across the 3.0 W resistor. V2 = I2R2:    V2 = (2.4)3 --> V2 = 7.2 volts

  9. The current through the battery (I):   I = I2/3 + I3  --> I = 2.0 + 2.4  = 4.4 amps