Which of the circuits shown above are wired in parallel? 1) A only 2) B only 3) C only 4) B and C 5) all of them

5 A 8 A 2 A P What is the current at point P? A)2 A B)3 A C)6 A D)10 A E) need more info to answer

Kirchhoff’s Rules ( rules of thumb ) 1. The SUM OF CURRENTS entering any junction equals the sum of all currents leaving. I1I1 I2I2 I3I3 I 1 = I 2 + I 3 IBIB IAIA ICIC I C = I A + I B This follows from Conservation of Charge!

24 volts + - When fully charged at Q=CV=(12pF)(24v) the circuit has two competing voltage steps: +24v – 24v = 0

12  6  12 V The voltage drop is 1.greatest across R1. 2.greatest across R2. 3.greatest across R3. 4. the same across each of the three resistors. 3  R1 R2 R3

2  4  12 V The current through the 2  resistor is 1.half the current through the 4 . 2.the same as thru the 4 . 3.twice the current through the 4 . 1.½ Ampere2. 1 Ampere 3.2 Ampere 4. 3 Ampere 5. 4 Ampere 6. 6 Ampere I

2  4  –IR = –8 V–IR = – 4 V +  = 12 V Start + 12 V– 8 V– 4 V = 0  V = Current in each resistor? Total resistance = 6  Ohm’s Law: I= V / R = 12 V / 6  = 2 A Voltage drop across a resistor? V = I R 12 V I

Kirchhoff’s Rules ( rules of thumb ) 2. The SUM OF POTENTIAL DIFFERENCES across all components of ANY closed circuit loop is ZERO. This follows from Conservation of Energy!  R1R1 R2R2 R3R3 R4R4  = IR 1 + IR 2 + IR 3 + IR 4 or  - IR 1 - IR 2 - IR 3 - IR 4 = 0

Voltage drops enter with a – sign and voltage gains enter with a + sign in this equation.  I R + + – – –IR +IR

–I 2 R 1 –I 2 R 2 ++ Start +  – I 2 R 1 – I 2 R 2  V = Loop Rule Example (without numbers) = 0 + I 3 R 3 –– Start –  + I 3 R 3  V = = 0 The first thing we have to do is? Define currents! I1I1 I2I2 I3I3 Now apply Loop rule to each loop R1R1 R2R2 R3R3

Circuits The light bulbs in the circuit are identical. When the switch is closed both bulbs go out A) both bulbs go out the intensity of both bulbs increases B) the intensity of both bulbs increases the intensity of both bulb decreases C) the intensity of both bulb decreases D) nothing changes

Circuits The light bulbs in the circuit are identical. When the switch is closed, the intensity of bulb A increases a) the intensity of bulb A increases the intensity of bulb A decreases b) the intensity of bulb A decreases the intensity of bulb B increases c) the intensity of bulb B increases d) the intensity of bulb B decreases e) nothing changes

1) A only When resistors are in parallel, the voltage drop across each is the same. C)6 A 3. greatest across R3. In series circuits the same current surges through each resistor! The voltage drop across is IR. Same I, but different Rs! 2. the same as thru the 4  Ampere D) nothing changes e) nothing changes See discussion on following slide

R 12 V R I +12V +12V - I 0 R - I 0 R = 0 I 0 = 12V/R Write a loop law for original loop: Write a loop law for the new loop: +12V - I 1 R = 0 I 1 = 12V/R The key here is to determine the potential (V a –V b ) before the switch is closed. From symmetry, (V a –V b ) = +12V. Therefore, when the switch is closed, NO additional current will flow! Therefore, the current after the switch is closed is equal to the current before the switch is closed....or R 12 V R I a bAnswer