Momentum and electricity problems and solutions
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The document discusses power and resistance in electric circuits. It provides formulas for calculating power (P), resistance (R) in series and parallel circuits, and uses these to solve sample circuit problems. Key points are: - Power (P) equals current (I) squared times resistance (R). - Resistance in series adds. Resistance in parallel is calculated using reciprocal formula. - Sample circuit is solved step-by-step using the series/parallel resistance formulas and Ohm's law.
Momentum and electricity problems and solutions
- 2. POWER IN ELECTRIC CIRCUITS 푃 = 퐼푉 푃 훼 퐼, 푉 푐표푛푠푡푎푛푡, 푒푠푝푒푐푖푎푙푙푦 푖푛 푝푎푟푎푙푙푒푙 푐푖푟푐푢푖푡푠 ℎ푒푛푐푒: 푡ℎ푒 푙표푤푒푟 푟푒푠푖푠푡표푟 풃풖풓풏풔 풃풓풊품풉풕풆풓 P = 퐼2푅 푃 훼 푅, 퐼 푐표푛푠푡푎푛푡 푖푛 푠푒푟푖푒푠 푐푖푟푐푢푖푡 , HENCE, LOWER RESISTOR BURNS DIM 푃 = 푉2 푅 (푃 훼 푉2), 푅 푐표푛푠푡푎푛푡
- 3. PROBLEM 65 CHAPTER 20 Information to use • When two or more resistors are in series, the equivalent resistance is given by Rs R1 R2 R3 . . . • When resistors are in parallel, the expression to be solved to find the equivalent resistance is given by 1 Rp 1 R1 1 R2 1 R3 ... . We will successively apply these to the individual resistors in the figure in the text beginning with the resistors on the right side of the figure. • Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance of the combination of those two resistors is 10.0 . • The 9.0- and 8.0- resistors are in parallel; their equivalent resistance is 4.24 . • The equivalent resistances of the parallel combination (9.0 and 8.0 ) and the series combination (4.0 and the 6.0 ) are in parallel; equivalent resistance is 2.98 . . • The 2.98- combination is in series with the 3.0- resistor, so that equivalent resistance is 5.98 . • Finally, the 5.98- combination and the 20.0- resistor are in parallel, so the equivalent resistance between the points A and B is 4.6 Related problem no 64 page 632
- 4. PROBLEM 85 CHAPTER 20 STEP 1: CHOOSE A JUNCTION AND DIRECTIONS OF CURRENTS AND ALLOCATE SIGNS ON RESISTORS STEP 2: APPLY THE JUNCTION RULE : 푰ퟏ + 푰ퟑ = 푰ퟐ … … … … … … (ퟏ) STEP 3: APPLYING THE LOOP RULE LOOP ABCD CLOCKWISE: −ퟐ 푰ퟏ + ퟔ푽 + ퟒ 푰ퟑ + ퟑ푽 = ퟎ … … … … … ퟐ LOOP BEFC CLOCKWISE: −ퟖ 푰ퟐ − ퟗ 푽 − ퟒ 푰ퟑ − ퟔ푽 = ퟎ … … … … … ퟑ SOLVING SIMULTANEOUS EQUATIONS WE GET 푰ퟑ = −ퟏ. ퟖퟐ 푨 SIGNIFICANCE OF MINUS SIGN: + B + + + + 3.00 V 9.00 V C 6.00 V A D E F 2.00 Ω 8.00 Ω I1 I2 I3 4.00 Ω + The minus sign indicates that the current in the 4.00- resistor is directed downward , rather than upward as selected arbitrarily in the drawing.
- 5. CHAPTER 7: PROBLEM 25
- 6. REASONING • No net external force acts on the plate parallel to the floor; therefore, the component of the momentum of the plate that is parallel to the floor is conserved as the plate breaks and flies apart. • Initially, the total momentum parallel to the floor is zero. After the collision with the floor, the component of the total momentum parallel to the floor must remain zero. • The drawing in the text shows the pieces in the plane parallel to the floor just after the collision. • Clearly, the linear momentum in the plane parallel to the floor has two components; therefore the linear momentum of the plate must be conserved in each of these two mutually perpendicular directions. • Using the drawing in the text, with the positive directions taken to be up and to the right, we have m v m v 1 1 2 2 (sin 25.0 ) + (cos 45.0 ) = 0 (1) y direction m v m v m v 1 1 2 2 3 3 (cos 25.0) + (sin 45.0) – = 0 (2) Substituting known values and solving equations simultaneously we get m1 1.00 kg m2 1.00 kg . Related example 9 section 7.4 in your prescribed book: Cutnell and Johnsons 9th Edition