Forproblemsinvolvingequivalentresistance涉及的等效电阻的问题.doc

AP Physics: Analysis of DC CIRCUITS3 of 3For problems involving equivalent resistance:1. Determine whether the resistors are arranged in series or in parallel.2. If necessary, simplify the circuit step by step until it is reduced to a simple series or parallel combination. Use the appropriate formula to determine the equivalent resistance.3. Use Ohms law and Kirchhoffs laws to solve for the current through each resistor and the potential difference across each resistor.For problems involving a complex circuit (Kirchhoffs Laws):1. Assign a direction to the current in each branch of the circuit. Place a positive sign on the side of each resistor where the current enters and a negative sign where the current exits.2. Place a positive sign at the positive terminal of each battery and a negative sign at the negative terminal.3. Select a junction point and use Kirchhoffs junction rule to write an equation for the currents entering the point. 4. Apply Kirchhoffs loop rule and write equations based on the gains and losses of potential around selected loops.5. Solve the equations for the unknown currents algebraically.Problems1. Find the equivalent resistance of the parallel resistors:It is clear that these 4 are all in parallel, so use 1/R = 1/R1 + 1/R2 1/R = 1/2 + 1/3 + 1/6 + 1/1 (finding common denom) = 3/6 + 2/6 + 1/6 + 6/6 = 12/6so R = 1/(12/6) = W NOTE: the resistance is LOWER than the smallest resistor in the parallel group. 2. What fraction of the potential would be lost across the 5W resistor and across the parallel resistors?V5 = 5/5 Vparallel = /5. NOTE: the voltage drop across any parallel path will be the same!3. Find the total resistance. Since the 5 is in series with those in parallel, we simply add: 5 + = 5 W 4. If the ideal ammeter reads 3.20 A, what will the ideal voltmeter read? Hint: all of these are in parallel. Find the voltage across the 4W first.Since these are all in parallel, the voltage drop across each will be the same.V = IR = 4ohms*3.2A = 12.8VThe current through the top would be 12.8V/21ohmsThe current through the bottom would be 12.8V/14ohms5. Find the current and voltage loss in each resistor and the voltage in the source if the ammeter reads 2.1A.There are 2 nodes, one before, and one after the parallel resistors (12ohm and 6 ohm). The current going into the 4ohm (call it I1)will be the same as that going through the 2ohm (call it I4) and the battery, but it will split through the 12 (I3) and 6 (I4). Since I3 must go through the 12ohm and the ammeter, the voltage drop for the parallel resistors is 12ohms*2.1 = 25.2V. The current through the 6ohm would be 25.2V/6 = 4.2A(we expect it to be twice as much since its resistance is of the 12ohm). Now that we know the current through both branches of the parallel, we know that the current through the 4ohm and the 2ohm will just be the sum of those = 2.1 + 4.2 = 6.3Amps.The Equiv. resistance would be 4 + 1/(1/12 + 1/6) + 2 = 10ohms. The potential for the circuit is therefore V = (6.3A)(10ohms) = 63V6. How much voltage is lost across each lamp? How much current passes through each lamp? How much power is used in each?The 3 and the 6 ohms are in parallel, so the equiv resistance for those 2 would be 1/(1/3 + 1/6) = 2ohms. The voltage lost across that will be 6V*(2/(4+2) = 2V. The voltage lost across the top lamp would be 6V*(4/(4+2) = 4V.The current through the 3ohm: 2V/3ohm = .6667amp P = .6667amp* 2V = 1.333W6ohm: 2V/6ohm = .3333A P =.3333A*2V = .6667W 4ohm: 4V/4ohm = 1A, but we know that must be true since the current must be .6667A + .3333A from the two branches and P = 1A*4V = 4W7. What is the equivalent resistance between points A and B of the circuit shown? How much power would be dissipated by this circuit if a constant 20-V dc source with a 0.10- internal resistance was placed across A and B? The 6, 1, 4, and 3 are in parallel with the 4, 8, and 8You get 6 + 2 for the upper branch and 4 + 4 for the lower8 in parallel with 8 gives 4ohms total resistanceIf there are also 0.10ohm from the battery, there is a tota