In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. (E = 8.50 V, r; = 10 kn, and r2 = 16 kn.) =10.0 µF 3.00 k2 Figure P28.67 (a) Find the steady-state current in each resistor. I = 327 I2 = 327 13-kn = 0 HA V HA HA (b) Find the charge Q on the capacitor. 52 (c) The switch is opened at t = 0. Write an equation for the current IR, in R2 as a function of time. O (327 HA)e-t/(0.190 s) O (275 µA)et/(0.190 s) O (275 µA)e-t/(0.190 s) O (327 µA)et/(0.190 s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value. ms

The emfs in the figure below are &₁ = 5.00 V and E2 = 18.0 V. The resistances are R₁ when the switch is in the following states. (a) open I1 = 0.188 I2 = 0.188 I3 = 0.188 14 = 0 I1 (b) closed R4 = I₂ = I3 = 14 A A A A A A X www R₁ R₂ ww R3 + E2 0.8285 Your response differs from the correct answer by more than 100%. A .106 Your response differs from the correct answer by more than 100%. A 18.00, R₂ = 30.5 , R3 = 47.50, and R4 58.0 . Find the magnitude of the current in each resistor

The resistor R in Figure P28.83 receives 20.0 W of power. Determine the value of R.

A 10 μF capacitor with an initial energy of 0.1 J is discharged through a 7.5 MΩ resistor. What is the initial charge on the capacitor? The charge, Q0 =  Units             . What is the current through the resistor when the discharge starts? The current, I0 =  Units             . Determine the potential difference across the capacitor and the rate at which the thermal energy is dissipating in the resistance 0.9 min after the discharge starts. The potential difference across C, VC =  Units             . The power dissipated in R, P =  Units             .