GENERAL PHYSICS 2

Please answer all of the Activities (1, 2 and 3) given below. Please provide the right answers with effective solutions. Please refer to the references given below also (right after the Activities image).

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ACTIVITIES

Activity 1 Problem Solving (Answer without soluaon will not be considered) 40 10 0 R1 R4 Solve for a. R2,3 b. R6,7 5 V R6 C. R5,6,7 + R2 20 40 d. R4, 5,6,7 R5 e. R2,3,4,5,6,7 40 f. Rtotal R7 R3 8 n 20 g. ItotalActivity 2 Problem Solving (Answer without solution will not R1 E1 be considered) MA Given: = = 12.00 v E2 = 9.00 v R2 R, = 2.000 MAA R2 = 5.000 R3 = 10.000 R3 E2 Activity 3 Essay (10 points) Note: Same grammar and exact content of answer with your classmates mean zero for the entire Activity sheet. Explain how the Motion of a Charged Particle in a Magnetic Field causes the northern light also known as aurora BorealisSubject Matter: Series and Parallel Connection, Kirchhoff's Loop and Junction POINTS TO REMEMBER Ohm's Law Relates the voltage across a resistor to the current flowing through it as a given by the equation V = IR voltage = (electric current) (Resistance) Combination of Resistors (100) Series and Parallel Circuits. Series Circuit . It is a circuit that contains more R. R2 than one electrical component Wv connected one after the other in a single path. a. Drawing diagram b. Schematic diagram (a) (b) Parallel circuit It is a circuit in which two or more electrical components are connected across each other in such a way that the current is Wv distributed between them. In such a circuit, each component RZ K2 WV operates independently of the others. a. Drawing diagram b. Schematic diagramResistors . The resistance of a resistor is called the equivalent resistance Rtotal of the combination Properties of Series and Parallel Circuit Quantity Series resistors Parallel resistors It= = 12=13= Current (I) h=hthat3+.. tin V. = Vi + V2 + V/3 Voltage (V) V. = VI=V2 = V3 = ". = Vn Equivalent 1 Resistance Rt = Ri + R2 + R3 1 1 1 (Rtotal) TR, R3 Example: Three resistors with values of 60.0 9, 30.0 $2 and 20.0 $2, respectively, are connected in series to a 110.0 V battery of negligible internal resistance. a. Draw a circuit diagram b. Find equivalent resistance of the combined resistors c. Current flowing through each resistor a. Schematic b. Total resistance diagram "Rt = R1 + R2 + R3 R. = 60.00 + 30.00 + 20.00 = 110.0n R2 E. Total current Vt 110.0 v it = = = 1.0 A Re x 110.0 0 Find the equivalent resistance of R2 the combination of resistors shown R Rs at the right side. RA Given: 9 00 R1 = 5.00 R2 = 3.50 R3 = 2.50 R4 = 24.00 Rs = 9.20Solution: Ra and R3 are in series. R23 = 3.50 + 2.50 = 6.00 However, R23 and R4 are in parallel. Their combined resistance R2,3,4 is computed as follows: 1 1 R2,3,4= 1 = 4.80 R23 RA 6.01 24.00 Replacing the parallel group of R2,3, and R4 Dy Im,-,. is computed as follows: Rt = R, + R2,34 + Rs - 5.0n + 4.80 + 9.20 = 19.0n Kirchhoff's Rule The following are the steps to follow when applying Kirchhoff's rule: 1. Label the current in each branch as It, I, and so on. 2. Assume a direction for each current 3. In writing the junction equation, the current entering a junction is positive, whereas the current leaving a junction is negative 4. In applying the loop rule, begin at a point in the loop and go around the loop in a clockwise or counter clockwise manner. This MODULE will follow the clockwise movement in each loop 5. In writing the loop equation. the following sign conventions must be observed: a. The electromotive force is positive if it traverses from the negative terminal to the positive terminal. It is negative if it traverses from the positive terminal to the negative terminal b. The potential difference across a resistor is negative when the resistor is traversed in the direction of the current; otherwise, it is positive. 6. The unknown currents are usually determined by solving simultaneous lop equations as well as junction equations. Example: Determine I1, Iz and Is given that ada atofales: to rolls radmos so 8 = 5.00 v 82 = 10.00 v 83 = 12.00 v R1 = 2.00 R2 = 3.00 R3 = 4.00From (1) 4 = 12 + 13 (6) Substituting (6) to (3) (-5.00V) + (12 + /3) (2.0) + 12(3.0) + (10.00 v) = 0 (-5.00V) + (2.082)/2 + (2.02)13 + (3.02)/2 + (10.00 v) = 0 (7) (5.000)125.00V 2.00 Substituting (7) for & in (4), (-10.00 v) -12 (30n) (120ev)+ (-5.00.0)12 - 5.00V 2.0n (4.0.) = 0 (-13.004)12 =320V $12 - 2.45A The negative sign of 12 means that the direction of Iz is opposite to that of the assumption made. However, the negative sign of /, must be retained involving of the other unknowns. Using (7) ( 5.000)(2.464) -SOOVA 13 $200 3.65A Using (6) 1 = 3.65A - 2.46A = 3.65A Motion of a Charged Particle in a Magnetic Field : A charged particle experiences a force when moving through a magnetic field. What happens if this field is uniform over the motion of the charged particle? What path does the particle follow? The simplest case occurs when a charged particle moves perpendicular to a uniform B-field. If the field is in a vacuum, the magnetic field is the dominant factor determining the motion. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The particle continues to follow this curved path until it forms a complete circle. Another way to look at this is that the magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle's kinetic energy and speed thus remain constant. The direction of motion is affected but not the speed.Solution: loop 3 RI The currents for each branch are labelled and a direction for each current is assumed. There are only two junctions in the given circuit, A E1 loop RZ and 8. the junction equation for A points A and B are as follows: loop 2 For junction A, 41-3 (1) For junction B, 1+121650 (2) Note that these equations are equivalent. The three loops that may be considered for the given circuit are shown. Starting with point A and going clockwise around each loop will yield the following loop equations E, .- 5.00 V R = 200 0 For loop 1, loop ). -5.001 + 1, (2.00) + 12(3.00) +10.070 E, 10.00 V . 3.00 1 A For loop 2, E? > 10.00 V -E2 - 12R2 - 83 + 13 R3 = 0 -10.00 v - 12(3.00) -12.00 D + 73(4.07) = 0 5, : 3.00 V For loop 3, (5) A loop -5.00V + 1, (2.0n) - 12.00 v + 13 (4.0/) = 0 RI = 4.00 n E, = 12.00 VFigure 8.3.1 A negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular to the paper (represented by the small s-like the tails of arrows). The magnetic force is Paper perpendicular to the velocity, so velocity changes in direction but not magnitude. The result is uniform circular motion. (Note that because the charge IS negative, the force is opposite in direction to the prediction of the right-hand rules