Problem%20Set%204%20Solutions

Problem%20Set%204%20Solutions - 6. The current in the...

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6. The current in the circuit is i = (150 V – 50 V)/(3.0 + 2.0 ) = 20 A. So from V Q + 150 V – (2.0 ) i = V P , we get V Q = 100 V + (2.0 )(20 A) –150 V = –10 V.
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10. (a) The work done by the battery relates to the potential energy change: ( ) 12.0V 12.0 eV. qV e V e ∆= = = (b) P = iV = neV = (3.40 × 10 18 /s)(1.60 × 10 –19 C)(12.0 V) = 6.53 W.
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18. (a) For each wire, R wire = ρ L/A where A = π r 2 . Consequently, we have R wire = (1.69 × 10 8 m Ω⋅ )(0.200 m)/ π (0.00100 m) 2 = 0.0011 . The total resistive load on the battery is therefore tot R = 2 R wire + R = 2( 0.0011 Ω) + 6.00 Ω = 6.0022 . Dividing this into the battery emf gives the current tot 12.0 V 1.9993 A 6.0022 i R ε == = . The voltage across the R = 6.00 resistor is therefore Vi R (1.9993 A)(6.00 ) = 11.996 V 12.0 V. (b) Similarly, we find the voltage-drop across each wire to be wire wire R (1.9993 A)(0.0011 ) = 2.15 mV. (c) P = i 2 R = (1.9993 A)(6.00 ) 2 = 23.98 W 24.0 W. (d) Similarly, we find the power dissipated in each wire to be 4.30 mW.
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( V A V B ) – V 3 = 78 36 = 42 V, which implies the current is i 1 = (42 V)/(2.0 ) = 21 A. By the junction rule, then, the current in R 2 = 4.0 is i 2 = i 1 i = 21 A 6.0 A = 15 A.
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This note was uploaded on 05/04/2010 for the course PHYS 11b taught by Professor Mobius during the Spring '09 term at Harvard.

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Problem%20Set%204%20Solutions - 6. The current in the...

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