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ps06sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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p. 1 of 10 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2010 Problem Set 6 Solutions Problem 1: Four Resistors Four resistors are connected to a battery as shown in the figure. The current in the battery is I , the battery emf is ε , and the resistor values are R 1 = R , R 2 = 2 R , R 3 = 4 R , R 4 = 3 R . (a) Rank the resistors according to the potential difference across them, from largest to smallest. Note any cases of equal potential differences. Resistors 2 and 3 can be combined (in series) to give 23 2 3 2 4 6 R R R R R R = + = + = . 23 R is in parallel with 4 R and the equivalent resistance 234 R is 23 4 234 23 4 (6 )(3 ) 2 6 3 R R R R R R R R R R = = = + + Since 234 R is in series with 1 R , the equivalent resistance of the whole circuit is 1234 1 234 2 3 R R R R R R = + = + = . In series, potential difference is shared in proportion to the resistance, so 1 R gets 1/3 of the battery voltage ( 1 /3 V ε Δ = ) and 234 R gets 2/3 of the battery voltage ( 234 2 /3 V ε Δ = ). This is the potential difference across 4 R ( 4 2 /3 V ε Δ = ), but 2 R and 3 R must share this voltage: 1/3 goes to 2 R ( 2 (1/3)(2 /3) 2 /9 V ε ε Δ = = ) and 2/3 to 3 R ( 3 (2/3)(2 /3) 4 /9 V ε ε Δ = = ) . The ranking by potential difference is 4 3 1 2 V V V V Δ > Δ > Δ > Δ . (b) Determine the potential difference across each resistor in terms of ε . As shown from the reasoning above, the potential differences are 1 2 3 4 2 4 2 , , , 3 9 9 3 V V V V ε ε ε ε Δ = Δ = Δ = Δ = (c) Rank the resistors according to the current in them, from largest to smallest. Note any cases of equal currents.
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p. 2 of 10 All the current goes through 1 R , so it gets the most ( 1 I I = ). The current then splits at the parallel combination. 4 R gets more than half, because the resistance in that branch is less than in the other branch. 2 R and 3 R have equal currents because they are in series. The ranking by current is 1 4 2 3 I I I I > > = . (d) Determine the current in each resistor in terms of I . 1 R has a current of I . Because the resistance of 2 R and 3 R in series ( 23 2 3 2 4 6 R R R R R R = + = + = ) is twice that of 4 3 R R = , twice as much current goes through 4 R as through 2 R and 3 R . The current through the resistors are 1 2 3 4 2 , , 3 3 I I I I I I I = = = = (e) If R 3 is increased, what happens to the current in each of the resistors? Since 23 4 2 3 4 1234 1 234 1 1 23 4 2 3 4 ( ) R R R R R R R R R R R R R R R + = + = + = + + + + increasing 3 R increases the equivalent resistance of the entire circuit. The current in the circuit, which is the current through 1 R , decreases. This decreases the potential difference across 1 R , increasing the potential difference across the parallel combination. With a larger potential difference the current through 4 R is increased. With more current going through 4 R , and less in the circuit to start with, the current through 2 R and 3 R must decrease. Thus, 4 1 2 3 increases and , , and decrease I I I I (f) In the limit that R 3 , what are the new values of the current in each resistor in terms of I , the original current in the battery?
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