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Unformatted text preview:  1  Physics 2213 HW #8 — Solutions Fall 2008 25.81 [Internal Resistance] The terminal voltage is ab V Ir IR ε == , where R is the resistance connected to the battery. During the charging the terminal voltage is ab V Ir = + . P VI = and energy is E Pt = . 2 I r is the rate at which energy is dissipated in the internal resistance of the battery. (a) 12.0 V (10.0 A) (0.24 ) 14.4 V. ab V Ir = + = + Ω = (b) 6 (10 A) (14.4 V) (5) (3600 s) 2.59 10 J. E Pt IVt = = = = × (c) 2 2 5 diss diss (10 A) (0.24 ) (5) (3600 s) 4.32 10 J. E P t I rt = = = Ω = × (d) Discharged at 10 A: 12.0 V (10 A) (0.24 ) 0.96 . 10 A Ir I R r R IΩ = ⇒ = = = Ω + E E (e) 6 (10 A) (9.6 V) (5) (3600 s) 1.73 10 J. E Pt IVt = = = = × (f) Since the current through the internal resistance is the same as before, there is the same energy dissipated as in (c): 5 diss 4.32 10 J. E = × (g) Part of the energy originally supplied was stored in the battery and part was lost in the internal resistance. So the stored energy was less than what was supplied during charging. Then when discharging, even more energy is lost in the internal resistance, and only what is left is dissipated by the external resistor. 26.13 [Four Light Bulbs] In both circuits, with and without R 4 , replace series and parallel combinations of resistors by their equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and voltages in the original circuit. Use 2 P I R = to calculate the power dissipated in each bulb. (a) The circuit is sketched below. 2 3 4 , , and R R R are in parallel, so their equivalent resistance eq R is given by eq 2 3 4 1 1 1 1 R R R R = + + eq eq 1 3 and 1.50 . 4.50 R R = = Ω Ω The equivalent circuit: ( 29 1 eq I R R+ = E 1 eq I R R = + E 1 9.00 V 1.50 A and 1.50 A 4.50 1.50 I I = = = Ω + Ω Then ( 29 ( 29 1 1 1 1.50 A 4.50 6.75 V V I R = = Ω = ( 29 ( 29 eq eq eq eq 1.50 A, 1.50 A 1.50 2.25 V I V I R = = = Ω = For resistors in parallel the voltages are equal and are the same as the voltage across the equivalent resistor, so 2 3 4 2.25 V. V V V = = = 2 3 4 2 3 4 2 3 4 2.25 V 0.500 A, 0.500 A, 0.500 A 4.50 V V V I I I R R R = = = = = = = Ω 2  Note that 2 3 4 1.50 A, I I I + + = which is eq . I For resistors in parallel the currents add and their sum is the current through the equivalent resistor....
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This note was uploaded on 10/25/2008 for the course PHYS 2213 taught by Professor Perelstein,m during the Fall '07 term at Cornell.
 Fall '07
 PERELSTEIN,M
 Magnetism, Resistance, Energy, Heat

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