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KIRCHHOFF’S LAWS 1. Voltage Rule 2. Current Rule 1. Voltage Rule. This is based on the conservation of energy: “the sum of voltages around a closed conducting loop (that is, a circuit) must be zero” Since voltage and work are related, we are saying the net work done must be zero. Observe the diagram below: Resistor R1 Supplies energy for work Voltage does work (uses energy) + Net Work = 0 The Resistor can’t do more work than is available (it can’t do less either - if there’s too much energy generated, it burns out) 2. Current Rule. This is based on the conservation of charge: “the sum of currents entering a junction = the sum leaving” [junction: a place where 2 or more conductors join] I So we see that I1 I2 Resistor R1 I = I1 + I2 This relationship is the essence of the Current Rule How does this apply to Series and Parallel Circuits? SERIES RESISTANCE Let’s examine a Series Circuit First…. Resistor R1 Resistor R2 1.5 Volts + 1.5 Volts + Current, I: one path start here: see below 1st: current rule = since there are no junctions this means that the Current, I, never changes. So we only have one current. 2nd: voltage rule = the voltage inputs and decreases must add up to zero! To do this, one must make a trip around the circuit counting voltage inputs and voltage drops. You can go clockwise or CCW but you MUST maintain that direction in the loop. Going Clockwise, we’ll start at the lower left corner. We add any voltage inputs and any voltage drops: +1.5 V + 1.5 V - V (lost to resistor 1) - V (lost to resistor 2) = 0 We can use OHM’S LAW, V = IR, to replace “V” in both terms above with “IR” +3.0 V - IR1 - IR2 = 0 Note that batteries in series add their voltages!!! So we have: +3.0 V = IR1 + IR2 +3.0 V = I (R1 + R2) +3.0 V = I (Requivalent) Therefore, Requivalent = R1 + R2 + R3 + --- Rn for series resistance Example: Problem: a) Find the “equivalent” resistance of the circuit shown b) Find the current in the circuit shown c) Find the voltage (energy) loss thru each resistor a) Just add up all the resistances! Rseries = R1 + R2 + R3 = 60 + 100 + 20 = 180 ohms This is called the “Equivalent Circuit” (because the power used is equivalent to the original, complex circuit) b) Now it’s easy to find the Current; just use OHM’S LAW: V = IR 9 Volts = I (180 ohms) I = 0.05 amperes c) Now, since the current, I, is the same for all the resistors, then, V(across resistor 3) = IR3 V3 = (0.05 amps)(20 ohms) V3 = 1 volt answer Do the same for the other resistors. See if you get voltage drops totaling 9 volts across all the resistors… PARALLEL RESISTANCE Note that the voltage is the same for all resistances!!! (however, the work won’t be the same because it depends on the current flow…recall W = -qV …so it depends on how much charge is moving) From the Current Rule we know that I = I1 + I2 Thus I V V 1 1 V( ) R1 R2 R1 R2 I V( 1 REquivalent ) So that in general, 1 1 1 1 1 ... R1 R2 R3 Rn Requivalent For Parallel Resistors Or, in a more useful format: Requivalent 1 ( R11 R12 R13 ... R1n ) Example: Problem: Find a) the equivalent resistance for this circuit Solution: Requivalent 1 ( R11 R12 ) 1 1 1 Requivalent 1 16.67 1 ( 100 20 ) 0.01 0.05 0.06 Notice that this is lower than either of the actual resistors. b) Find the current I: Solution: This result leaves us with an “equivalent” circuit: For which we can easily find the general current using Ohm’s Law: V = IR V = IR I = V/R = 3.0 volts/(16.67 ) = 0.18 amperes c)Find the currents I1 and I2 !! Solution: Finding the individual currents in this basic parallel circuit is very straightforward once we realize that ALL the resistors have 3.0 Volts across them! Therefore, using Ohm’s Law: V = IR we find… through the 100 Ohm resistor: through the 20 Ohm resistor: I = V/R = 3.0 / 100 = 0.03 amps I = V/R = 3.0 / 20 = 0.15 amps Example 2: What happens if we put another battery in the circuit? Now it becomes a little more difficult to determine the current…we have to use “loop” analysis (based on Kirchhoff’s voltage rule: the voltage around a closed loop must add up to zero). Let’s see how that works: we have: If we follow the arrow around we see we increase by 3.0 volts and then pass through the 100 ohm resistor, which is going to use up voltage. How much? Since V = IR, it will use I1(100) volts. So 3.0 - I1(100) = 0 or I1 = 3/100 = 0.03 amps going through the 100 ohm resistor. Notice this didn’t change from the previous example. Now for the second loop: We follow this loop around and see that 1. we go up 3.0 volts again 2. we pass through the 20 ohm resistor and use up voltage equal to Vused = I2(20) 3. We go BACKWARDS through the 1.2 volt battery meaning…? These must all cancel out to zero, so… 3 - I2(20) - 1.2 = 0 1.8 - I2(20) = 0 - I2(20) = -1.8 I2 = 1.8/20 = 0.09 amps So I1 = 0.03 amps I2 = 0.09 amps What does I = ??? Notice it’s lower than our first example because we’ve put in an opposing battery.