HW1-Solutions-enme350

HW1-Solutions-enme350 - Page 4 of 4 Problem 8 In the...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ENME 350 Electronics & Instrumentation I Homework #1 - Solutions Page 1 of 4 Problem 1 Solution Problem 2 Solution A node is a point that joints two or more circuit elements. All points joined by ideal conductors are electrically equivalent. Thus, there are five nodes in this circuit at hand. Note that node 1 and node 5 are trivial nodes . Node 5
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
ENME 350 Electronics & Instrumentation I Homework #1 - Solutions Page 2 of 4 Problem 3 Solution Problem 4 Solution Problem 5 Solution
Background image of page 2
ENME 350 Electronics & Instrumentation I Homework #1 - Solutions Page 3 of 4 Problem 6 Use Kirchhoff’s current law to determine the unknown currents I 2 and I 3 in the circuit of Figure P6. Assume that I 0 = -2 A , I 1 = -4 A , I S = 8 A . Figure P6. Solution Applying KCL to node (a) and node (b): I 0 + I 1 + I 2 = 0 I 0 + I S + I 1 ± I 3 = 0 I 2 = - ( I 0 + I 1 ) = 6 A I 3 = I 0 + I S + I 1 = 2 A Problem 7 Apply KVL to find the voltages 1 v and 2 v in Figure P7. Figure P7. Solution Applying KVL: V v v V v v 12 0 10 3 5 2 0 3 5 1 1 2 2 = = + - + - = = + + -
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
ENME 350 Electronics & Instrumentation I Homework #1 - Solutions
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Page 4 of 4 Problem 8 In the circuit shown in Figure P8, determine the terminal voltage of the source, the power supplied to the circuit (or load), and the efficiency of the circuit. Assume that the only loss is due to the internal resistance of the source. Efficiency is defined as the ratio of load power to source power. Figure P8. Solution : 0, , (using Ohm's law) 12 1 mA 12 7 1 mA 7 . S S T T T L T S S T L T S T S T T L T KVL V R I V V R I V R I R I V V I R R k V R I k V-+ + = = ∴-+ + = ⇒ = = = + Ω ⇒ = = Ω× = ( ) 2 2 2 3 7 7 mW 7 10 12 V 1 mA 12 mW R T L L L VS S T V V V P R R P V I = = = = × Ω = -= -× = -Recall that the negative sign for the source power means that the source is supplying power. Therefore, the supplied power is 12 mW. The efficiency is given by 7 mW 0.5833 58.33% 12 mW L S R out in V P P or P P η = = = = . Ω = Ω = = k 7 k 5 V 12 L S S R R V...
View Full Document

This note was uploaded on 02/27/2010 for the course ENME 271 taught by Professor Marks during the Spring '08 term at Maryland.

Page1 / 4

HW1-Solutions-enme350 - Page 4 of 4 Problem 8 In the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online