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SUPERPOSITION%20fixed - V 4 = 10V 4(2 4 6 8 12 = 1.25 V...

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SP1 SUPERPOSITION Superposition is the method used to solve circuits with multiple sources. Sources are taken individually and the results added/subtracted. PROCEDURE (VOLTAGE SOURCES ONLY)- 1. Short all voltage sources except one. 2. Solve for current thru a particular resistor (given). 3. Short this voltage source and unshort another. 4. Repeat steps 2 and 3 until all sources have been analyzed. 5. Add (+) or subtract (-) currents based on flow direction. 6. Voltage drop = IR. EXAMPLE 1 – Find V 2 . If sources are side by side (no resistor in between): V 2 = (V 1 – V 2 ) * [R 2 / (R 1 + R 2 + R 3 + R 4 )] (Voltage Division)
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SP2 EXAMPLE 2 – Find V 4 . STEP 1) Short 5V source:
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Unformatted text preview: V 4 = 10V * [ 4 / (2 + 4 + 6 + 8 + 12) ] = 1.25 V STEP 2) Short 10V source: V 4 = 5V * [ 4 / (2 + 4 + 6 + 8 + 12) ] = 0.625 V STEP 3) Superimpose results: Current in opposite directions => 1.25V – 0.625V = 0.625 V SP3 EXAMPLE 3 – Find V 6 . SP4 IF CURRENT SOURCES ARE PRESENT – To use superposition with current sources, open (do not short) them. Find V 2 . STEP 1) Open (remove) current source. V 2 = 10 * [ 2 / (2 + 4 + 6 + 8) ] = 1 V STEP 2) Short voltage source. I L (shown) = 2A * [ (4 + 6) / (2 + 8+4+6) ] = 1A V 2 = 1A * 2Ω = 2V STEP 3) Current directions opposite => 2V – 1V = 1V = V 2...
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