# Capacitor Analysis: Step-by-Step Summary **Given:** - Capacitance: \( C = 132\,\mu\text{F} = 132 \times 10^{-6}\,\text{F} \) --- ## (a) Find \( i_c(t) \) from \( v_c(t) = 7.2 + 19.4\cos^2(165t)\,\text{V} \) ### 1. Simplify \( v_c(t) \) using double-angle identity: \[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \] \[ v_c(t) = 7.2 + 19.4 \left( \frac{1 + \cos(330t)}{2} \right ) = 16.9 + 9.7\cos(330t) \] ### 2. Differentiate to find current: \[ i_c(t) = C \frac{dv_c}{dt} = C \cdot [-9.7 \cdot 330 \sin(330t)] \] \[ i_c(t) = -132 \times 10^{-6} \times 9.7 \times 330 \sin(330t) \] \[ i_c(t) = -.4225 \sin(330t)\,\text{A} =422.5\sin(330t)\,\text{mA} \] **Final Answer:** \[ \boxed{i_c(t) = -422.5 \sin(330t)\, \text{mA}} \] --- ## (b) Maximum Stored Energy - Maximum voltage: \( v_{max} = 16.9 + 9.7 = 26.6\,\text{V} \) \[ w_{max} = \frac{1}{2} C v_{max}^2 = \frac{1}{2} \times 132 \times 10^{-6} \times (26.6)^2 \] \[ w_{max} = .0467\,\text{J} = 46{,}699\,\mu\text{J} \] **Final Answer:** \[ \boxed{w_{max} = 46{,}699\,\mu\text{J}} \] --- ## (c) Find \( v_c(t) \) from \( i_c(t) = 19.4 e^{-165t}\,\text{mA},\ v_c()= \) ### 1. Express current in amperes: \[ i_c(t) = 19.4 \times 10^{-3} e^{-165t}\,\text{A} \] ### 2. Integrate: \[ i_c(t) = C \frac{dv_c}{dt} \implies dv_c(t) = \frac{i_c(t)}{C}dt \] \[ v_c(t) = \int_^t \frac{19.4 \times 10^{-3} e^{-165\tau}}{132 \times 10^{-6}} d\tau \] \[ = \frac{19.4 \times 10^{-3}}{132 \times 10^{-6} \cdot 165} (1 - e^{-165t}) \] \[ = .8907(1 - e^{-165t})\,\text{V} = 890.7(1 - e^{-165t})\,\text{mV} \] **Final Answer:** \[ \boxed{v_c(t) = 890.7(1 - e^{-165t})\,\text{mV}} \] --- ## (d) \( v_c() = 7.2\,\text{mV} \) Initial Value \[ v_c(t) = v_c() + 890.7 (1 - e^{-165t})\,\text{mV} = 7.2 + 890.7(1 - e^{-165t})\,\text{mV} \] **Final Answer:** \[ \boxed{v_c(t) = 7.2 + 890.7 (1 - e^{-165t})\,\text{mV}} \] --- ## (e) Given Stored Energy \( w(t) = 17e^{-560t}\,\mu\text{J} \) ### 1. Express in joules: \[ w(t) = 17 \times 10^{-6} e^{-560t}\,\text{J} \] ### 2. Find \( v_c(t) \): \[ w(t) = \frac{1}{2} C v_c^2(t) \implies v_c(t) = \sqrt{\frac{2w(t)}{C}} \] \[ = \sqrt{\frac{2 \times 17 \times 10^{-6} e^{-560t}}{132 \times 10^{-6}}} = .5075 e^{-280t}\,\text{V} \] ### 3. Find \( i_c(t) \): \[ i_c(t) = C \frac{dv_c}{dt} = C \cdot \left(-.5075 \times 280 e^{-280t} \right) \] \[ = -132 \times 10^{-6} \times .5075 \times 280 e^{-280t} = -18.76 \times 10^{-3} e^{-280t}\,\text{A} = -18.76 e^{-280t}\,\text{mA} \] **Final Answers:** \[ \boxed{i_c(t) = -18.76 e^{-280t}\,\text{mA}} \] \[ \boxed{v_c(t) = .5075 e^{-280t}\,\text{V}} \] --- # 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 calculations assume SI unit consistency and standard calculus for capacitor charge/energy relationships. No diagrams are included as all steps are algebraic.