140 likes | 415 Views
Series Circuits:. Other examples:. have only 1 path for current. Series circuits - ________________________________________ _________________________________________. could have switches, etc. I. circuit element ___. 1. wire. wire. high. ________ potential. +. voltage
E N D
Series Circuits: Other examples:
have only 1 path for current Series circuits - ________________________________________ _________________________________________ could have switches, etc. I circuit element ___ 1 wire wire high ________ potential + voltage source circuit element ___ 2 - low ________ potential wire wire circuit element ___ 3 I Assume: 1. _____________________________________________________ 2. _____________________________________________________ 3. _____________________________________________________ 4. _____________________________________________________ All energy provided by source is used in the elements Wires have no potential drop (voltage) across them Pos. current is out of the “high” voltage side No charge is “lost.” All current returns to the source.
For a circuit with 2 resistors: I V1 V I1 R1 I2 R2 V2 Energy V1 + V2 _______________ Conservation: V = Charge _______________ Conservation: I = I1 = I2 Equivalent R1 + R2 _______________ (Total) Resistance: Req = I Req Ohm’s __________ Law applies to the total: V = and to each individual element: V1 = V2 = I1R1 I2R2
V = IR Ohm’s Law I = V/R
Ex. Find all the voltages and currents in the circuit below: 40W 20. V 120W I = I1 = I2 V1 = I1R1 40W 5.0 V V1 = I1 = R1 = 0.125 A V2 = I2R2 120W 15.0 V V2 = I2 = R2 = 0.125 A 20/160 A 0.125 A 160W 20. V V = I = Req = Req = R1 + R2 I = V/Req
ratio Form the _________ of each resistance to Req = _______ , and then multiply by the __________ voltage V 160 W total V = 20. V R1 40W 40W x 20 V = 5 V Req 160W 120W R2 120W 15 V x 20 V = 160W Req • V “divides up” ______________________________ as the R’s • This is because ___________ R requires _________ energy. • Series circuits are _______________________________. in the same proportion more more voltage dividers
+ back to ____ side of the battery Plot V vs. “distance around circuit.” ____ side of battery + wire 20 V dropped across the ______ resistor 40 W 15 wire potential difference (V) V dropped across the ______ R - at the ___ side of the battery 160 W wire 0 distance around circuit R = 0 No ______ drop across wires because we assume _________
Important: “I is ______________ everywhere in ___________ circuit” does NOT mean that I is ___________ in _________________ circuit! a series the same every other the same I= I1= I2= 10/100 = .1 A 25W R1= 10. V .1 A 75W R2 = .1 A I= I1= I2= I3 = 10/200 = .05 A R1= 100W 25W R2= .05 A 10. V .05 A 75W R3= .05 A same I is still the _______________ in all parts of the second circuit, but it is a ________________ I than the first one! different
If you replace the resistors of a circuit with one resistor, the total I would be the same Equivalent resistance: _________________________________ ________________________________________________________ Replacing this part of the circuit with a single _______________ resistor: Req = R1 + R2 = = 20. V 40W equivalent 40 W + 120 W 120W 160 W …gives you this circuit: V Req The total I = = 20 V 160 W 20. V Req = 160 W = 0.125 A same I This is the ____________ as before.
series All _______________ circuits can be ___________________ in this way. simplified A. Req = _____W V = 20 V V = 20 V 50W 200 150W B. Req = _____W V = 12 V 8W 24 16W V = 12 V voltage source This can be done even if the ______________________ is not shown. C. Req = _____W 70W 90 20W Req = _____W 10W D. 5W 50 15W 20W original same I Req results in the _____________ as the _________________ circuit.
Meter __________ Hookups: Original circuit: R1 V R2 disconnect To measure I1, the current through R1, _________________ the circuit and _____________ an ________________ next to R1 ammeter insert insert disconnect A R1 R1 V V R2 R2 Other possibilities: R1 V V R1 R2 R2 A A anywhere gives same I. ___ is the same everywhere, so _________________________ I
R1 R1 R1 V V V R2 R2 R2 R1 R1 V V R2 R2 do NOT To measure, V1, the voltage across R1, __________disconnect the circuit. Simply connect the ______________ across R1 voltmeter Original circuit: V Other possibilities: V R1 V V R2 total Similarly, to measure the _________ voltage V or V2: V V