HW1-Solutions-enme350

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

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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
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ENME 350 Electronics & Instrumentation I Homework #1 - Solutions Page 2 of 4 Problem 3 Solution Problem 4 Solution Problem 5 Solution
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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 = = + - + - = = + + -
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ENME 350 Electronics & Instrumentation I Homework #1 - Solutions
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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...
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