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Basic circuit analysis

Electrical and Electronic Principles I

Course

Electrical and Electronic Principles I (MBE202)

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City University of Hong Kong

Academic year: 2017/2018

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City University of Hong Kong

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Oct. 23, 2018 Test 1 Total 100 Marks (IERG2060) (90 minutes) Total 125 Marks (ESTR2304) Notes: 1. An cheat sheet is allowed. 2. Only calculator is allowed. 3. Answer ALL five (5) questions. 4. Put your answers on this question paper. 5. Use ballpoint pens. NO PENCIL. Q1: Fig. 1. (a) Determine Req in Fig. 1. (b) Determine the power supplied the voltage source. (c) (For ESTR2304 only) If the voltage of the voltage source is doubled, what will happen to its power supplied? Answer: (a) ( 40 )( 40 ) 20 ( 22 17) 40 40 40 40 60 ( )( 24 40 60 40 60 Q1 Answer ( 44 )( 44 ) 44 ( 20 24 ) 44 44 10 (18 22 ) eq (10 )( 10 40 (b) P V 2 102 12 W 8 Req (c) ( ) 2 new Req 202 W 4 Pold 8 If voltage is doubled, power is quadrupled. Q2 Answer (b) 70 110 ( ) ( ) 11k 252k 192k 421k 120 0 25 110 0 15k 36k ( ) 0 ( ) 187 15 120 70 0 2 25 210k 6k 0 2 ( ) 0 276 15 120 70 0 3 110 25 0 10k 0 ( ) 24k 206 15 i2 i3 i1 187 445 mA 421k 276 656 mA 421k 206 490 mA 421k (c) 25 i2 25 ( 656 ) 16 W 15 i3 15 ( 490 ) 7 W Q3: Fig. 3. (a) Find the Thevenin equivalent to the left of terminals (i., not including RL). (b) Find the Norton equivalent to the left of terminals (i., not including RL). (c) Find RL that will draw the maximum power. (d) Determine that maximum power. (e) (For ESTR2304 only) If RL is only half of the value found in (c) and it consumes a power of px, what is another value of RL that will consume the same power of px? Answer: (a) 160 I eq 800 mA 200 100 200 200 eq Veq 800m )(100 ) 80 V Req 100 100 200 I eq 80 400 mA 200 Q4: Fig. 4. The switch in Fig. 4 has been closed for a long time. At t 0, the switch is opened. (a) Determine the initial capacitor voltage v(0). (b) Determine the time constant (c) Calculate v(t) for t 0. (d) What is v when t 2 s? (e) (For ESTR2304 only) Find the capacitor current i(t) for t 0? Answer: (a) v ( 0) 10 (16 ) 4 V 30 10 (b) eq C (10 ) 1 ms (c) v ( t ) v ( 0 ) e t 4e t V (d) v(2) 4e 2) (Note: The voltage will already be less than one percent after 5 time constants. After 2 seconds, i., time constants, the voltage must be 0.) (e) dv ( t ) dt )( ) e i (t ) C t mA Q4 Answer Q5 Answer (f) ) 5 4 ( t1 5 A v ( t1 ) V OF TEST

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