problem26_68

problem26_68 - The 00 . 8 and the 00 . 6 resistors are now...

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26.68: a) Ignoring the capacitor for the moment, the equivalent resistance of the two parallel resistors is = = + = 00 . 2 ; 00 . 6 3 00 . 3 1 00 . 6 1 1 eq eq R R In the absence of the capacitor, the total current in the circuit (the current through the 00 . 8 resistor) would be A 20 . 4 2.00 8.00 V 0 . 42 = + = = R i ε of which 3 2 , or 2.80 A, would go through the 00 . 3 resistor and 3 1 , or 1.40 A, would go through the 00 . 6 resistor. Since the current through the capacitor is given by , RC t e R V i - = at the instant 0 = t the circuit behaves as through the capacitor were not present, so the currents through the various resistors are as calculated above. b) Once the capacitor is fully charged, no current flows through that part of the circuit.
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Unformatted text preview: The 00 . 8 and the 00 . 6 resistors are now in series, and the current through them is A. 3.00 ) 6.00 (8.00 / V) . 42 ( = + = = R i The voltage drop across both the 00 . 6 resistor and the capacitor is thus V. . 18 ) 00 . 6 ( ) A 00 . 3 ( = = = iR V (There is no current through the 00 . 3 resistor and so no voltage drop across it.) The change on the capacitor is C 10 7.2 V) (18.0 farad) 10 00 . 4 ( 5 6-- = = = CV Q...
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