# Step-by-Step Summary and Final Answers Given: Capacitance, \( C = 132\ \mu F = 132 \times 10^{-6}\ \mathrm{F} \) --- ## (a) Find the capacitor current \( i_c(t) \) given \( v_c(t) = 7.2 + 19.4 \cos^2(165t)\ \mathrm{V} \) ### **Steps:** 1. **Expand \( \cos^2 \) term:** \( \cos^2(\theta) = \frac{1+\cos(2\theta)}{2} \) So, \[ v_c(t) = 7.2 + 19.4 \left( \frac{1 + \cos(330t)}{2} \right) = 7.2 + 9.7 + 9.7 \cos(330t) = 16.9 + 9.7 \cos(330t) \] 2. **Differentiate to find current:** \[ i_c(t) = C \frac{dv_c}{dt} = C \cdot [-9.7 \cdot 330 \sin(330t)] \] \[ i_c(t) = -132 \times 10^{-6} \times 9.7 \times 330 \sin(330t) = -.4225 \sin(330t)\ \mathrm{A} = -422.5 \sin(330t)\ \mathrm{mA} \] ### **Answer:** \[ \boxed{i_c(t) = -422.5\sin(330t)\ \mathrm{mA}} \] --- ## (b) Maximum stored energy in the capacitor ### **Steps:** 1. **Maximum voltage:** \( v_{max} = 16.9 + 9.7 = 26.6\ \mathrm{V} \) 2. **Energy formula:** \[ w_{max} = \frac{1}{2} C (v_{max})^2 = \frac{1}{2} (132 \times 10^{-6}) (26.6)^2 = .0467\ \mathrm{J} = 46{,}699\ \mu \mathrm{J} \] ### **Answer:** \[ \boxed{w_{max} = 46{,}699\ \mu \mathrm{J}} \] --- ## (c) Find \( v_c(t) \) given \( i_c(t) = 19.4 e^{-165t}\ \mathrm{mA} \), \( v_c() = \) ### **Steps:** 1. **Rewrite units:** \( i_c(t) = 19.4 \times 10^{-3} e^{-165t}\ \mathrm{A} \) 2. **Integrate:** \[ i_c(t) = C \frac{dv_c}{dt} \implies dv_c(t) = \frac{i_c(t)}{C}dt \] \[ v_c(t) = \int_^t \frac{19.4 \times 10^{-3} e^{-165\tau}}{132 \times 10^{-6}} d\tau = \frac{19.4 \times 10^{-3}}{132 \times 10^{-6} \times 165}(1 - e^{-165t}) \] \[ = .8907(1 - e^{-165t})\ \mathrm{V} = 890.7(1 - e^{-165t})\ \mathrm{mV} \] ### **Answer:** \[ \boxed{v_c(t) = 890.7(1 - e^{-165t})\ \mathrm{mV}} \] --- ## (d) Find \( v_c(t) \) if \( v_c() = 7.2\ \mathrm{mV} \), same \( i_c(t) \) as (c) ### **Steps:** \[ v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ \mathrm{mV} \] ### **Answer:** \[ \boxed{v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ \mathrm{mV}} \] --- ## (e) Given \( w(t) = 17e^{-560t}\ \mu \mathrm{J} \), find \( v_c(t) \) and \( i_c(t) \) ### **Steps:** 1. **Convert energy to Joules:** \( w(t) = 17 \times 10^{-6} e^{-560t}\ \mathrm{J} \) 2. **Voltage from energy:** \[ w(t) = \frac{1}{2} C v_c^2(t) \implies v_c(t) = \sqrt{\frac{2w(t)}{C}} = \sqrt{ \frac{2 \cdot 17 \times 10^{-6} e^{-560t} }{ 132 \times 10^{-6} } } = .5075 e^{-280t}\ \mathrm{V} \] 3. **Current:** \[ i_c(t) = C \frac{dv_c}{dt} = C \left[ -.5075 \cdot 280 e^{-280t} \right] = -132 \times 10^{-6} \times .5075 \times 280 e^{-280t} = -18.76 \times 10^{-3} e^{-280t}\ \mathrm{A} = -18.76 e^{-280t}\ \mathrm{mA} \] ### **Answers:** \[ \boxed{i_c(t) = -18.76 e^{-280t}\ \mathrm{mA}} \] \[ \boxed{v_c(t) = .5075 e^{-280t}\ \mathrm{V}} \] --- ## Final Summary Table | Part | Expression | Units | |------|------------------------------------------------------|------------| | (a) | \( i_c(t) = -422.5\sin(330t) \) | mA | | (b) | \( w_{max} = 46{,}699 \) | μJ | | (c) | \( v_c(t) = 890.7(1 - e^{-165t}) \) | mV | | (d) | \( v_c(t) = 7.2 + 890.7(1 - e^{-165t}) \) | mV | | (e) | \( i_c(t) = -18.76e^{-280t} \), \( v_c(t)=.5075e^{-280t} \) | mA, V | --- **Note:** All answers are boxed for clarity and units are stated. No steps or information omitted from your summary.