Model the battery and the connecting wires are ideal

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Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure 32.38a. Solve: After the switch closes at t = 0 s, the capacitor begins to charge. At time t , let the current and the charge in the circuit be i and q , respectively. Also, assume clockwise direction for the current i . Using Kirchhoff’s loop law and starting clockwise from the lower left corner of the loop, 0 q dq q iR R C dt C + = = + E E RC dq = ( C E – q ) dt dq dt C q RC = E Integrating both sides, ( ) ( ) ( ) 0 0 0 ln ln ln Q t Q dq dt t t C q C Q C C q RC RC RC = ⇒ − = ⇒ − + = E E E E ( ) ln 1 t RC t RC C Q t C Q e Q C e C RC C = − = = E E E E E Letting Q max = C E and τ = RC , we get Q = Q max (1 e t/ τ ).

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32.78. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure 32.38a. Solve: (a) According to Equation 32.36, during charging the charge on the capacitor increases according to ( ) max 1 t Q Q e τ = . Therefore, the current in the circuit behaves as ( ) max 1 t t RC t RC dQ C I Q e e e dt RC R τ τ = = = = E E Using Equation 32.8, the power supplied by the battery as the capacitor is being charged is 2 bat t RC t RC P I e e R R = = = E E E E Because bat P dU dt = , we have 2 2 2 bat bat 0 0 0 t RC t RC dU P dt dU P dt e dt RC e C R R = = = = = E E E That is, the total energy which has been supplied by the battery when the capacitor is fully charged is E 2 C . (b) The power dissipated by the resistor as the capacitor is being charged is 2 2 2 resistor 2 t RC P I R R e R = = E Because resistor P dU dt = , we have dU = P resistor dt 2 2 2 2 2 resistor 0 0 2 2 t RC t RC RC C dU e dt U e R R = = = E E E (c) The energy stored in the capacitor when it is fully charged is 2 2 2 2 max C 1 1 1 2 2 2 Q C U C C C = = = E E (d) For energy conservation, the energy delivered by battery is equal to the energy dissipated by the resistor R plus the energy stored in the capacitor C. This is indeed the case because U bat = E 2 C , 2 1 resistor 2 U C = E , and 2 1 C 2 U C = E .
32.79. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure CP32.79. Solve: (a) During charging, when the neon gas behaves like an insulator, the charge on the capacitor increases according to Equation 32.36. That is, Q = Q max (1 e t/ τ ). Because Q = C Δ V C , Δ V C = Δ V C0 (1 e t/ τ ) = E (1 e t/ τ ) Let us say that the period of oscillation begins when Δ V = V off and it ends when Δ V = V on . Then, ( ) off off off off off off 1 ln t t V t V e e V τ τ τ = = = E E E E E Because the period T = t on t off , we have off off on off on on ln ln ln ln V V T RC V V V V τ τ τ = = = E E E E E E E E (b) Substituting the given values into the above expression and noting that and 1 0.10 s T f = = , ( ) ( ) 6 6 90 V 20 V 0.10 s 0.10 s 10 10 F ln 5.1 k 90 V 80 V 10 10 F ln7 R R = × = = Ω ×

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• Winter '10
• E.Salik
• Resistor, Potential difference, Ω, Series and parallel circuits

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