Download Week1_Solutions

EE215 Class Problems, Week 1 Solutions All questions are initially listed in black. After class, those that were covered in class using Top Hat (as class activities) will be highlighted in blue. Questions remaining in black (after class) may be used for additional review. Solutions are noted in italics. Week 1, Q1 In the following circuit, i = 5 cos (π*t) A and v = 100 sin (π*t)) V. The cosine and sine terms are given in radians (not degrees). a . At t = 0.75 sec, is power delivered to or extracted from Element B? at t = 0.75 sec; i = 5 cos (3 π/4) = 5 * (-­‐0.71) = -­‐3.5 A and v = 100 sin (3 π/4) = 100 (0.71) = 71 V current is flowing in the direction of voltage rise, so power is being extracted. b. How much power in Watts is delivered or extracted in (a)? Give your answer as a positive quantity. Power = V*I = 3.5 * 71 = 249 Watts c. Assume Element B is a passive element (resistor, capacitor, or inductor). Which single passive element is Element B most likely to be? Since the derivative of sin (x) is cos (x); the current is a function of the derivative of voltage in this situation. Therefore i = C dv/dt; the element must be a capacitor. d. What is the value of the element you determined in part (c)? Include correct units. 5 cos (π*t) A = C (d/dt) 100 sin (π*t) = C * π * 100 * cos (π*t) Solving gives C = 16mF Week 1, Q2 Find the current i in mA using current divider: Current Divider can only be applied to two resistors in series at a time. So, we must first combine the two 10,000 ohm resistors in parallel to get 5,000 ohms. Then, we can apply current divider: i = 5mA * 5,000 / (24,000 + 6,000 + 2,000 + 5,000) = 0.676 mA Week 1, Q3 Find the current i in mA using current divider: The inductor acts like a short circuit (with a resistance of zero) under DC steady state conditions. Thus, it shorts the other two branches of the circuit out, causing the current i to go to 0A. Week 1, Q4 Find the current i in mA using current divider:
The capacitor acts as an open circuit under steady state conditions, so it is essentially not there (not doing anything in the circuit). Thus, the current i is simply: 5mA * (10,000/(10,000 + 32,000)) = 1.19 mA Week 1, Q5 Find the voltage v using voltage divider: The voltage across the 10,000 ohm resistor AND the 5,000 ohm, 2,000 ohm, and 3,000 ohm resistors in combination is 10V (because voltages in parallel must be equal). Therefore, v = 10* (2,000/10,000) = 2V Week 1, Q6 Find the voltage v using voltage divider: The capacitor acts as an open circuit under DC steady state conditions, preventing any current from reaching the resistors in the circuit. Therefore v must equal 0. Week 1, Q7 Find the voltage v using voltage divider: The inductor acts as a short circuit under DC steady state conditions. Therefore, the circuit is identical to question 5 and v is 2V. Week 1, Q8 Is the following circuit valid? (Hint: Use KVL and Ohm’s Law) KVL would require 20V + Vcurrent_source -­‐ 20* 3A = 0 (note that the polarity in the resistor must be + on the right and – on the left, because the 3A is flowing clockwise; we have assumed the current source has a voltage rise in the direction of current flow). Vcurrent_source =60 -­‐ 20 = 40V This is entirely possible, so yes, the circuit is valid Week 1, Q9 Is the following circuit valid? KCL requires that: 3mA + -­‐0.001vx = 0; thus vx = 3V On the right side of the circuit, KVL requires that 3 – vx = 0; 3 -­‐3 = 0 Therefore, this circuit is valid; note that if the 3V source were anything but 3V, the circuit would no longer be valid. Week 1, Q10 Is the following circuit valid if the resistor R1 = 100 ohms? KCL requires that: 0.1A + 0.1v = 0; therefore v must be equal to -­‐1V KVL requires that v – (0.1*R1) – Vvccs = 0; and -­‐1 – 0.1*R1 -­‐ Vvccs = 0 There is no reason that this should not work, so yes, the circuit is valid Week 1, Q11 A 9V battery is connected to a 1MOhm resistor. The voltage across the 1MOhm resistor is 8.8V. If you were to model the 9V battery as an ideal voltage source, what value would it have? It would be an 8.8V ideal voltage source. The 1MOhm resistor is large enough that the internal resistance of the battery can be ignored. Week 1, Q12 A 9V battery is connected to a 1MOhm resistor. The voltage across the 1MOhm resistor is 8.8V. When connected to a 100 ohm resistor, the voltage drops to 8.2V. What is the value of the internal resistance of the battery in Ohms? 8.2V = (100/ (100 + Rs)) * 8.8V 0.932 (100 + Rs) = 100 Rs = 7.3 Ohms Week 1, Q13 A voltage divider circuit is wired onto a breadboard with a 9V battery placed in series with a 10 ohm resistor and a 50 ohm resistor. When measured with a multimeter, the values of the resistors are 9 ohms and 51 ohms. When measured with a multimeter, the value of the battery is 9V (exactly). What is the expected output voltage if measured between the 51 ohm and 9 ohm resistors if the 51 ohm resistor is connected on one end to the negative terminal of the battery and the 9 ohm resistor is connected on one end to the positive terminal of the battery? You can ignore the internal resistance of the battery. Output voltage = 9V * 51/(51 + 9) =7.65V