Practice Problems - ECE 110 M-C Brunet easy 1 challenging 1...

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ECE 110 M.-C. Brunet Practice Problems Collection from all Transparencies M.-C. BRUNET ECE 110 UIUC 1.16 Compute the variation of charge from 0 to 7 ms for the following current: I = 5A 6 10 8 5 I(mA) T (ms) SOLUTIONS: 1. (a) 35mC (b) 27 μ C 2. (a) I = 0.16A (b) I = 0.64A I? (b) 10 18 protons/second are traveling to the right, and 3.10 18 electrons/second are traveling to the left. 1 2 (a) (b) 10 18 protons per second are traveling to the right. (a) What is the value of I if: easy challenging M.-C. BRUNET ECE 110 UIUC 2.16 E2 E3 E1 v + - m v 3 + - + - 5v v 2 + - Am im i1 i3 i2 = 3A i0 1 2 3 1 5 4 2 6 7 8 Name the pairs of elements in series, and in parallel. Give im, i3, v 3 , v m and i 0 im=3A, i 3 = 0, v 3 = 0 (= voltage across the ammeter). E3 is short-circuited (and im = i 2) v m=5v, i o=0 (voltmeter) (1,2) in series, (3,4) and (7,8) in parallel none 3 2 1 3 3 5 2 4 1 Name the pairs of elements in series, and in parallel. SOLUTIONS : (Ammeter) (Voltmeter)
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ECE 110 M.-C. Brunet Practice Problems Collection from all Transparencies M.-C. BRUNET ECE 110 UIUC 3.14 1 SOLUTIONS : 4 2 1 Compute the power dissipated in the resistor. 3 Ω 15V 2 Find the current i 12W + 10V - LOAD POWER SUPPLY i 3 Assuming a 15 kW electric appliance, how much power needs to be supplied by a generator with 80% efficiency? 4 The energy available from a generator is 1kWh. How long will a 5W lamp operate, assuming 100% efficiency? 3 P SRS = 75W i = -1.2A 18.75kW 200 hours M.-C. BRUNET ECE 110 UIUC 3.16 5 4 Ω 20 sin (2 π t) Compute the average power dissipated for both circuits in the 4 Ohm resistor. SOLUTIONS : 5 (a) P average = 50W (b) P average = 50W too! 4 Ω 20 sin (2 π t) (a) (b) 2 Ω
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ECE 110 M.-C. Brunet Practice Problems Collection from all Transparencies M.-C. BRUNET ECE 110 UIUC 3 4.16 2 SOLUTIONS : 2 i 1 = 2A, i 2 = -3A 4 1 3 v 1 = -6V, v 2 = 2V P 1 = -16W, P 2 = 32W, P 3 = 24W, P 4 = 48W, P 5 = -88W Find i 1 and i 2 Compute the power (in SRS) for each element. Check that they add up to zero! i = 12.15mA (If you find 12.5mA, you are not using the basic laws correctly) Find v 1 and v 2 1 4 3 2 1 5A 3A - + 7V 5V + - 4 2 5 3 1 + - v 1 v 2 + - + - 6V 4 Find the current i . + - v=.7 25V i 2k Ω 1 4 5 2 3 + - 2V - + 3V + - 6V 8A 8A M.-C. BRUNET ECE 110 UIUC 5.15 10 Ω 3 Ω 3 Ω 4 Ω 2 Ω 5 Ω 2 Ω 5 Ω 2 Ω 4 Ω a b Req 6 Ω 3 Ω Solutions : Req = 13 Ω Req = 858/215 Ω 3 8 Ω 3 2 Ω 3 Ω 1 Ω 4 ab Req (between a and b) = ? 2 1 1 2 3 Req = 21/16 Ω a b Req
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ECE 110 M.-C. Brunet Practice Problems Collection from all Transparencies M.-C. BRUNET ECE 110 UIUC 5.16 Solutions : 1 2 3 1 + - 2 k Ω 15V + - 7V i Compute i and find which elements are loads. 2 12 Ω 5A + - V 1 Compute V 1 3 1 Ω 8V 1V i 1 5A + - + - i 2 2 Ω i 3 Compute i 1 , i 2 , i 3 i = 4mA; loads are the 2 k Ω resistor and the 7V ideal voltage ‘source’ (gets charged). V 1 is not equal to zero! V 1 = 60 volts. (This is a common mistake) i 1 = 7A (voltage across 1 Ohm resistor is 7volts by KVL) , i 2 = 0.5A (use KVL 1 st ), i 3 = 11.5A (use KCL) M.-C. BRUNET ECE 110 UIUC 5.23 2 Give the current read by the (ideal) ammeter: (use CDR) 3 10A 5 Ω 7 Ω 10 Ω 3 Ω A M I M 25V 5 k Ω 7 Ω + - 6 k Ω 2 k Ω + v - + - V Use VDR to find the voltage V.
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This note was uploaded on 09/16/2009 for the course ECE 110 taught by Professor Haken during the Spring '08 term at University of Illinois at Urbana–Champaign.

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Practice Problems - ECE 110 M-C Brunet easy 1 challenging 1...

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