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Unformatted text preview: 261DIRECTCURRENT CIRCUITSThese solutions sometimes use Epsilon to denote the EMF of a battery (the curly, script E in the book). Sometimes it looks like Ε, sometimesε, soWATCH OUT. 26.55.IDENTIFY:We need to do series/parallel reduction to solve this circuit. SET UP:P=ε2R, where Ris the equivalent resistance of the network. For resistors in series, Rs= R1+ R2, and for resistors in parallel 1/Rp= 1/R1+ 1/R2. EXECUTE: R=ε2P=(48.0 V)2295 W=7.810 Ω. R12=R1+R2=8.00 Ω. R=R123+R4. R123=R−R4=7.810 Ω−3.00 Ω=4.810 Ω. 1R12+1R3=1R123. 1R3=1R123−1R12=R12−R123R123R12. R3=R123R12R12−R123=(4.810 Ω)(8.00 Ω)8.00 Ω−4.810 Ω=12.1 Ω. EVALUATE:The resistance R3is greater than R, since the equivalent parallel resistance is less than any of the resistors in parallel. 26.59. IDENTIFY:The terminal voltage of the battery depends on the current through it and therefore on the equivalent resistance connected to it. The power delivered to each bulb is , where I is the current through it. SET UP:The terminal voltage of the source is . EXECUTE:(a) The equivalent resistance of the two bulbs isThis equivalent resistance is in series with the internal resistance of the source, so the current through the battery is and the current through each bulb is 2.2 A. The voltage applied to each bulb is . Therefore, . (b) If one bulb burns out, then . The current through the remaining bulb is 2.9 A, and . The remaining bulb is brighter than before, because it is consuming more power. EVALUATE:In Example 26.2 the internal resistance of the source is negligible and the brightness of the remaining bulb doesn’t change when one burns out. 26.63. IDENTIFY:Apply the junction rule to express the currents through the resistors in terms of Apply the loop rule to three loops to get three equations in the three unknown currents. SET UP:The circuit is sketched in Figure 26.61....
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This note was uploaded on 12/18/2011 for the course PHYS 2B 2b taught by Professor Hirsch during the Spring '10 term at UCSD.
 Spring '10
 Hirsch

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