PHYSICS 120 : ELECTRICITY AND MAGNETISM TUTORIAL QUESTIONS

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1 DC CIRCUITS Question 61 PHYSICS 120 : ELECTRICITY AND MAGNETISM TUTORIAL QUESTIONS A 1000Ω 2.0W resistor is needed but only 1000Ω 1.0W resistors are available. How can the required resistance and power rating be obtained by a combinationof the available units? What power is then dissipated in each resistor? The current through the resistor can be found by P = VI = I 2 R = V 2 so the current through the 2.0W resistor would be P 2.0 I = R = 1000 = A The maximum current that can flow through the 1.0W resistor is 1.0 I = 1000 = A 1 hence the 1.0W resistors must be wired in parallel. For a maximum current of 10 A in the 5 circuit, two resistors in parallel each would carry < A, so this is permissible. 10 Two 1000 Ω resistors in parallel would result in an equivalent resistance of R R EQ = = 500Ω so two combinations like this in series results in an equivalent resistance of 1000Ω. The power dissipated in each resistor is ( ) 2 1 P = I 2 R = = 0.50W 5 Question 62 Each of the three resistors in the figure has a resistance of 2.0Ω and can dissipate a maximum of 18 W without becoming excessively heated. What maximum power can the circuit dissipate? page 1 of 6

2 The maximum current that can flow in the circuit is P 18 I = R = 2.0 = 3.0A since this is the current flowing in the single 2.0Ω resistor. The equivalent resistance for the whole circuit is R EQ = = 3.0Ω Hence the maximum power dissipated is P = I 2 R = = 27W Question 63 (a) Find the potential difference V ad in the circuit (b) Find the potential difference across the 4.00 V cell. (c) A battery of emf 17.0V and internal resistance 1.00Ω is inserted in the circuit at d with its positive terminal connected to the positive terminal of the 8.00 V battery. Find V bc between the terminals of the 4.00V battery. (a) Choosing a direction for the current as anti-clockwise and a direction for the loop as dcbad, Kirchhoff s second rule gives 9I 0.5I +4 6I 8I I = 0 This leads to a current of I = = 0.50A To find V ad, start at point a and move in the direction of the current to point d: V a 8I I = V d The potential difference V ad = V d V a is then V ad = V d V a = 8 (8 0.5) ( ) = 3.75V (b) The potential difference V cb is found by starting at point c and moving to point d: V c 0.5I +4 = V b V cb = = 3.75V page 2 of 6

3 (c) With a new battery inserted in the circuit, the Kirchhoff II equation becomes 9I 0.5I +4 6I 8I I 17 (1 I) = 0 This now leads to a current of The potential difference V cb is then I = 5 25 = 0.20A V cb = 4 (0.5 I) = 4 (0.5 ( 0.20)) = 4.10V Question 64 A dry cell having an emf of 1.55V and an internal resistance of Ω supplies current to a 2.00Ω resistor. (a) Determine the current in the circuit. (b) Calculate the terminal voltage of the cell. (a) The current in the circuit is given by 1.55 = ( ) I I = = 0.745A (b) The terminal voltage is the voltage drop across the 2.00 Ω resistor: terminal voltage = = 1.49V Question 65 How many cells, each having an emf of 1.5V and an internal resistance of 0.50Ω must be connected in series to supply a current of 5 A to operate an instrument having a resistance 3 of 6.0Ω? page 3 of 6

4 Supposetherearencells, thenthecurrentinthecircuit,thetotalemfandthetotalresistance is given by 1.5n = (0.5n+6.0) n = ( ) 5 6 = = 15 Question 66 A battery has an internal resistance of 0.50Ω. A number of identical light bulbs, each with a resistance of 15 Ω, are connected in parallel across the battery terminals. The terminal voltage of the battery is observed to be one-half the emf of the battery. How many bulbs are connected? Suppose there are n light bulbs. The total resistance for n light bulbs in parallel is 1 R EQ = = n 15 R EQ = 15 n The terminal voltage V is V = IR EQ = 1 2 E where E = (R EQ +0.50)I. Hence IR EQ = 1 2 (R EQ +0.50)I 1 2 R EQ = 1 2 (0.5) R EQ = 0.5 n = = 30 page 4 of 6

5 Question 68 Find the magnitude and direction of the current in the 2.0Ω resistor in the diagram. Using Kirchhoff s rules, the equations needed to find the currents are: KI at a : I 1 +I 2 +I 3 = 0 KII for loop abcda : 2I I 3 = 0 KII for loop aefba : I I 2 = 0 (I) gives : I 3 = I 1 I 2 substitute (IV) into (II) : 2I 2 5+3( I 1 I 2 ) = 0 3I 1 5I 2 5 = 0 multiply (III) by 3 : 3I 1 +6I 2 +3 = 0 (V) - (VI) : 11I 2 8 = 0 I 2 = 8 11 A Hence I 2 = 0.73A is the current through the 2.0Ω resistor and it flows from b to a. (I) (II) (III) (IV) (V) (VI) Question 69 Determine the voltage across the 5.0Ω resistor in the circuit below. Which end of the resistor is at the higher potential? The Kirchhoff equations are KI at a : I 1 +I 2 = I 3 KII for loop abcda : 10I I 3 = 0 KII for loop adefa : 10I I 1 = 0 (I) (II) (III) page 5 of 6

6 These are solved simultaneously as follows: rearrange (I) : I 2 = I 3 I 1 substitute (IV) into (II) : 10(I 3 I 1 ) 13+10I 3 = 0 The voltage drop across the 5.0Ω resistor is 20I 3 10I 1 13 = 0 multiply (III) by 2 : 20I 3 10I = 0 (IV) (V) (VI) (V) + (VI) : 20I 1 +3 = 0 I 1 = 3 20 A (VII) V = I 1 R = = 3 4 = 0.75V Current I 1 is positive, therefore it flows in the direction of f to a. That means that point a is at a lower potential than point f. page 6 of 6

= (0.400 A) (4.80 V) = 1.92 W = (0.400 A) (7.20 V) = 2.88 W

= (0.400 A) (4.80 V) = 1.92 W = (0.400 A) (7.20 V) = 2.88 W Physics 2220 Module 06 Homework 0. What are the magnitude and direction of the current in the 8 Ω resister in the figure? Assume the current is moving clockwise. Then use Kirchhoff's second rule: 3.00