problem26_12

problem26_12 - 26.12: From Ohm's law, the voltage drop...

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Unformatted text preview: 26.12: From Ohm's law, the voltage drop across the 6.00 resistor is V = IR = (4.00 A)(6.00 ) = 24.0 V. The voltage drop across the 8.00 resistor is the same, since these two resistors are wired in parallel. The current through the 8.00 resistor is then I = V R = 24.0 V 8.00 = 3.00 A. The current through the 25.0 resistor is the sum of these two currents: 7.00 A. The voltage drop across the 25.0 resistor is V = IR = (7.00 A)( 25.0 ) = 175 V, and total voltage drop across the top branch of the circuit is 175 + 24.0 = 199 V, which is also the voltage drop across the 20.0 resistor. The current through the 20.0 resistor is then I = V R = 199 V 20 = 9.95 A. ...
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This homework help was uploaded on 04/19/2008 for the course PHYS 4 taught by Professor Stuart during the Spring '08 term at UCSB.

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