** Sorry guys I couldn't answer some questions during the review. Here's the complete solutions. Part of the solutions of the sample exam: 1. Circuit diagram at t = 0: +ve end of scope probe * * -ve end of scope probe * 5k(ohm) * +--www--+***** | | +--| |--+ fully charged Circuit diagram at t = infinity +ve end of scope probe * * -ve end of scope probe * 5k(ohm) * +--www--+***** | | +--| |--+ fully discharged (a) Observations: 0.5V/div Voltage of capacitor at t = infinity = 0. Analysis: We can conclude that the lowest line of the screen of the oscilloscope indicates the value 0. Voltage of capacitor at t = 0 = 8 div * (0.5V/div) = 4V (b) Observations: Voltage is decreasing. Max voltage = 4V. Analysis: Max current = Max voltage/resistance = 4/5k = 0.8mA (c) Observations: The voltage is dropping in the scope's screen. At t = 0, the voltage occupies 8 divisions. At t = infinity, the voltage occupies 0 divisions. 37.5% = 3/8. Analysis: At t = 2 division, the division of the voltage equals to 3. So the voltage remained is 3/8 or 37.5% of its original value. 2 div * (0.5s/div) = 1s. So it takes 1 second for the voltage to drop to 37.5% of its highest value. (d) Observations: v(t) = v_A + v_B*(e^(-t/RC)) (from the book) Analysis: Find v_A and v_B first. v(0) = v_A + v_B = 4V v(infinity) = v_A = 0V v_B = 4 v(t) = 4*(e^(-t/RC)) From part (c), we know that at 1s, voltage = 37.5% * Max_voltage 37.5% * Max_voltage = 4 * (e^(-1/RC)) 37.5% * 4 = 4 * (e^(-1/5k*C)) C = 2.05 * 10^(-4)F 2. C R_1 D R_2 +--www--+--www--oA | > = 9V < R_3=10k --- > | | +-------+-------oB E F R_1 = 1k(ohm) R_2 = 1k(ohm) R_3 = 10k(ohm) (a) Assumptions: (1) Assume that positive current goes clockwisely. V_A = V_D because A is an open circuit. V_F = V_B because B and F are on the same wire. So V_AB = V_DF. By voltage divider, V_DF = (R_3/(R_1 + R_3))*(-9) V_DF = (10/11)*(-9) V_DF = -8.2V So V_AB = -8.2V (b) C R_1 D R_2 C R_1 D +--www--+--www-----+A +--www--+ | > | | > = 9V < R_3=10k | ==> = < R_(2+3) --- > | --- > | | | | | +-------+----------+B +-------+ E F E F First I find out V_D. V_D will help me find out I_AB. The rightmost circuit is the equivalent of the leftmost circuit. 1 1 1 ------- = --- + --- R_(2+3) R_2 R_3 1 1 1 ------- = --- + --- R_(2+3) 1k 10k R_(2+3) = 909.09(ohm) By voltage divider, V_DF = (R_(2+3)/(R_(2+3)+R_1))*(-9) V_DF = -4.285V V_D = V_DF - V_F V_D = -4.285V - 0 V_D = -4.285V Now find I_AB. I_DA = I_AB V_D - V_A --------- = I_AB R_2 -4.285 - 0 I_AB = ------------ 1k I_AB = -4.285mA 5. (a) F (b) F ; In the US, the outlet supplies 120 root-mean-square voltage (c) F ; 47*10m = 470ms, not 0.47ms (d) T (e) F (f) F ; volt*amperes per second = watt per second (g) T (h) ? ; find that out in the book! 7. (a) All resistors are 1k(ohm) in the leftmost circuit +-www-+-www-+-o | | | R_X R_Y R_total o-+-www-+-www-+ ==> o-www---www-o ==> o-www-o | | +-www-+ | | +-www-+ I will first turn the leftmost circuit to the middle equivalent circuit. R_X is the equivalent resistor of the 4 resistors on the left. R_Y is the equivalent resistor of the 2 resistors on the right. 1 1 1 1 1 --- = ---- + ---- + ---- + ---- R_X 1k 1k 1k 1k R_X = 250(ohm) 1 1 1 --- = ---- + ---- R_Y 1k 1k R_Y = 500(ohm) To calculate R_total from R_X and R_Y R_total = R_X + R_Y R_total = 250 + 500 R_total = 750(ohm) (b) All capacitors are 10pF in the leftmost circuit C_X +-||-||-||-+ +-||-+ | | | | C_total o-+ +-o ==> o-+ +-o ==> o-||-o | | | C_Y| +-||-||-||-+ +-||-+ I will first turn the leftmost circuit to the middle equivalent circuit. C_X is the equivalent capacitor of the 3 capacitors at the top. C_Y is the equivalent capacitor of the 3 capacitors at the bottom. 1 1 1 1 --- = ----- + ----- + ----- C_X 10p 10p 10p C_X = 3.3pF 1 1 1 1 --- = ----- + ----- + ----- C_Y 10p 10p 10p C_Y = 3.3pF To calculate C_total from C_X and C_Y C_total = C_X + C_Y C_total = 3.3pF + 3.3pF C_total = 6.6pF