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Common 3 Problem Solutions and Notes

Common 3 Problem Solutions and Notes - Notes and Solved...

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Notes and Solved Problems for Common Exam 3 (Does not include Induction) 8. MULTI LOOP CIRCUITS Key concepts : Multi loop circuits of batteries and resistors: loops, branches and junctions should be distinguished. A general problem is to find currents insides every branch if the resistances and EMF’s of the batteries are known. Three rules apply: Branch rule: inside every branch the current is the same. The current may be different in different branches. Junction rule: the sum of all currents coming into the junction is equal to the sum of all currents leaving the junction. Loop rule: the sum of all potential differences in any complete walk through any closed loop of the circuit is equal to zero. The strategy for solving multi-loop circuit problems is the following: 1. Find all branches and enumerate currents. 2. Find all junctions and establish the relationships between the currents 3. Apply Kirchoff loop rules to the loops in the circuit. The total number of equations to be written should be equal to the total number of currents, which are unknown. Typical problems related to multi loop circuits: Problem 5. What is the current inside the resistor R 1 ? What is the power released in the resistor R 1 ? A. 1A, 27 W B. 2A, 13 W C. 3 A, 27 W D. 4 A, 12 W E. 5 A, 5 W Solution. While this is a multi loop circuit, the answer to that
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particular question can be found very easily. We see that the first loop contains the battery E 1 and only this resistor. Let us we write the Kirchoff loop rule for this particular loop. The current is assumed to be directed down inside R 1 , therefore we walk through the loop clockwise: E 1 -i 1 *R 1 =0 From which we figure out that i 1 =E 1 /R 1 =9/3=3 A. The power released in the resistor (energy rate dissipated in the resistor) is P=i 1 *V=i 1 *i 1 *R=i 1 2 *R=3*3*3=27 W. This trick works when you can find such a loop in the multi loop circuit which contains only one resistor and (possibly many) batteries. Then you can apply the Kirchoof loop rule to this particular loop and find the current inside the resistor immediately! Answer C. Problem 6. Order branches of bulbs by brightness, dimmest first: A. I, II, III B. II, III, I C. III, I, II D. III, II, I E. II, I, III Solution. The brightness of each branch is proportional to the total power released in the bulbs of the branch. Assume that the resistance of the bulbs is the same and it is equal to R. Then the total power of branch I is: P 1 = i 1 *V 1 =( i 1 ) 2 *R, the total power of branch II is: P 2 = ( i 2 ) 2 *R + ( i 2 ) 2 *R = 2*( i 2 ) 2 *R, and the total power of branch III is: P 3 = ( i 3 ) 2 *R + ( i 3 ) 2 *R + ( i 3 ) 2 *R =3*( i 3 ) 2 *R. Note that the power is expressed via the current and resistance here. What about the currents i 1 , i 2 , i 3 ? We can find currents easily by applying the Kirchoff loop equations to the loops which accounts for the battery and the particular branch.
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