# qsoln - MAE140 Linear Circuits Summer 1 2005 Quiz Solutions...

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Unformatted text preview: MAE140 Linear Circuits Summer 1 2005 Quiz Solutions. Wednesday June 29 Question 1: Mesh Analysis Part (i): Note that, because of the current source i isolated in mesh one, we have that current i 1 = i . Thus, considering the mesh equation for the second ( i 2 ) mesh, we have- 12 i 1 + (12 + 8 + 4) i 2- 4 i 3 = , 24 i 2- 4 i 3 = 12 i 1 = 12 i. Similarly, in the third ( i 3 ) mesh, we have- 3 i 1- 4 i 2 + (4 + 2 + 3) i 3 = ,- 4 i 2 + 9 i 3 = 3 i 1 . Thus, parenleftbigg 24- 4- 4 9 parenrightbiggparenleftbigg i 2 i 3 parenrightbigg = parenleftbigg 12 3 parenrightbigg × i. To demonstrate the solution, note that parenleftbigg 24- 4- 4 9 parenrightbiggparenleftbigg 3 i 5 3 i 5 parenrightbigg = parenleftbigg 12 3 parenrightbigg × i. Part (ii): The voltage v indicated on the diagram is the voltage across the 12 Ω resistor mea- sured from bottom to top plus that across the 3 Ω resistor measured in the same sense. That is, v = 12 × ( i 1- i 2 ) + 3 × ( i 1- i 3 ) = 12 2 5 i + 3 2 5 i = 6 i. Part (iii): Resistor 12Ω 8Ω 4Ω 3Ω 2Ω current 2 5 i 3 5 i 2 5 i 3 5 i power 48 25 i 2 72 25 i 2 12 25 i 2 18 25 i 2 Total power is thus...
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qsoln - MAE140 Linear Circuits Summer 1 2005 Quiz Solutions...

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