# Model assume ideal wires as the capacitors discharge

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Model: Assume ideal wires as the capacitors discharge through the two 1 k Ω resistors. Visualize: The circuit in Figure EX32.31 has an equivalent circuit with resistance R eq and capacitance C eq . Solve: The equivalent capacitance is C eq = 2 μ F + 2 μ F = 4 μ F, and the equivalent resistance is eq 1 1 1 1 k 1 k R = + Ω Ω R eq = 0.5 k Ω Thus, the time constant for the discharge of the capacitors is τ = R eq C eq = (0.5 k Ω )(4 μ F) = 2 × 10 3 s = 2 ms

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32.32. Model: The capacitor discharges through a resistor. Assume that the wires are ideal. Solve: The decay of the capacitor charge is given by the Equation 32.34: Q = Q 0 e t/ τ . The time constant is τ = RC = (1.0 × 10 3 Ω )(10 × 10 6 F) = 0.010 s The initial charge on the capacitor is Q 0 = 20 μ C and it decays to 10 μ C in time t . That is, / 0.010 s 10 C 10 C (20 C) ln (0.010 s)ln 2 6.9 ms 20 C 0.010 s 0.010 s t t t e t μ μ μ μ = = − ⇒ − = =
32.33. Model: The capacitor discharges through a resistor. Assume ideal wires. Visualize: The switch in the circuit in Figure EX32.33 is in position a. When the switch is in position b the circuit consists of a capacitor and a resistor. Solve: (a) The switch has been in position a for a long time. That means the capacitor is fully charged to a charge Q 0 = C Δ V = C E = (4 μ F)(9 V) = 36 μ C Immediately after the switch is moved to the b position, the charge on the capacitor is Q 0 = 36 μ C. The current through the resistor is R 0 9 V 0.36 A 25 V I R Δ = = = Ω Note that as soon as the switch is closed, the potential difference across the capacitor Δ V C appears across the 25 Ω resistor. (b) The charge Q 0 decays as Q = Q 0 e t/ τ , where τ = RC = (25 Ω )(4 μ F) = 100 μ s Thus, the charge is ( ) ( ) 50 s 100 s 0.5 36 C 36 C 22 C Q e e μ μ μ μ μ = = = The resistor current is ( ) 50 s 100 s 0 0.36 A 0.22 A t I I e e τ μ μ = = = (c) Likewise, at t = 200 μ s, the charge is Q = 4.9 μ C and the current is I = 49 mA.

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32.34. Model: A capacitor discharges through a resistor. Assume ideal wires. Solve: A capacitor initially charged to Q 0 decays as Q = Q 0 e t/RC . We wish to find R so that a 1.0 μ F capacitor will discharge to 10% of its initial value in 2.0 ms. That is, ( ) ( ) ( ) ( ) 3 2.0 ms R 1.0 F 0 0 6 2.0 10 s 0.10 ln 0.10 1.0 10 F Q Q e R μ × = = − × ( ) 3 6 2.0 10 s 0.87 k 1.0 10 F ln(0.10) R × = = Ω × Assess: A time constant of τ = RC = (870 Ω )(1.0 × 10 6 F) = 0.87 ms is reasonable.
32.35. Model: A capacitor discharges through a resistor. Assume ideal wires. Solve: The discharge current or the resistor current follows Equation 32.35: 0 . t RC I I e = We wish to find the capacitance C so that the resistor current will decrease to 25% of its initial value in 2.5 ms. That is, ( ) 2.5 ms 100 0 0 0.25 C I I e Ω = ( ) ( ) 3 2.5 10 s ln 0.25 100 C × = − Ω C = 18.0 μ F

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32.36. Visualize: Please refer to Figure P32.36. Solve: Bulbs D and E are in series, so the same current will go through both and make them equally bright (D = E). Bulbs B and C are in parallel, so they have the same potential difference across them. Because they are identical bulbs with equal resistances, they will have equal currents and be equally bright (B = C). Now the equivalent resistance of B +
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