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✅ **Final Answers (Step-by-step summary)** Given: Capacitance, ( C = 132\ \mu F = 132 \times 10^{-6}\ F ) --- ### **(a)** If [ v_c(t) = 7.2 + 19.4 \cos^2(165t)\ V ] then using (\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) ] Now, [ i_c(t) = C \frac{dv_c(t)}{dt} = C[-9.7(330)\sin(330t)] ] [ i_c(t) = -132\times10^{-6} \times 9.7 \times 330 \sin(330t) ] [ i_c(t) = -0.4225 \sin(330t)\ A = -422.5 \sin(330t)\ mA ] ✅ **Answer:** [ \boxed{i_c(t) = -422.5\sin(330t)\ \text{mA}} ] --- ### **(b)** Maximum stored energy: [ w_{max} = \frac{1}{2} C (v_{max})^2 ] where (v_{max} = 16.9 + 9.7 = 26.6\ V) [ w_{max} = \frac{1}{2}(132\times10^{-6})(26.6)^2 = 0.0467\ J = 46{,}699\ \mu J ] ✅ **Answer:** [ \boxed{w_{max} = 46{,}699\ \mu J} ] --- ### **(c)** If [ i_c(t) = 19.4 e^{-165t}\ mA,\ v_c(0)=0 ] then [ i_c(t) = C \frac{dv_c(t)}{dt} \implies dv_c(t) = \frac{i_c(t)}{C}dt ] Integrating: [ v_c(t) = \frac{19.4\times10^{-3}}{132\times10^{-6}}\frac{1}{165}(1 - e^{-165t}) ] [ v_c(t) = 0.8907(1 - e^{-165t})\ V = 890.7(1 - e^{-165t})\ mV ] ✅ **Answer:** [ \boxed{v_c(t) = 890.7(1 - e^{-165t})\ \text{mV}} ] --- ### **(d)** If same (i_c(t) = 19.4e^{-165t}\ mA) but (v_c(0)=7.2\ mV), then [ v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ mV ] ✅ **Answer:** [ \boxed{v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ \text{mV}} ] --- ### **(e)** If stored energy (w(t) = 17e^{-560t}\ \mu J) Convert to Joules: (17\times10^{-6} e^{-560t}) [ w(t) = \frac{1}{2} C v_c^2(t) \Rightarrow v_c(t) = \sqrt{\frac{2w(t)}{C}} ] [ v_c(t) = \sqrt{\frac{2(17\times10^{-6} e^{-560t})}{132\times10^{-6}}} = 0.5075 e^{-280t}\ V ] Now current: [ i_c(t) = C \frac{dv_c}{dt} = C(-0.5075)(280)e^{-280t} ] [ i_c(t) = -132\times10^{-6} \times 0.5075 \times 280 e^{-280t} = -18.76\times10^{-3} e^{-280t}\ A ] [ i_c(t) = -18.76 e^{-280t}\ mA ] ✅ **Answers:** [ \boxed{i_c(t) = -18.76 e^{-280t}\ \text{mA}} ] [ \boxed{v_c(t) = 0.5075 e^{-280t}\ \text{V}} ] --- ### ✅ **Final Summary** | 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)=0.5075e^{-280t}) | mA, V |

Question:

✅ **Final Answers (Step-by-step summary)** Given: Capacitance, ( C = 132\ \mu F = 132 \times 10^{-6}\ F ) --- ### **(a)** If [ v_c(t) = 7.2 + 19.4 \cos^2(165t)\ V ] then using (\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) ] Now, [ i_c(t) = C \frac{dv_c(t)}{dt} = C[-9.7(330)\sin(330t)] ] [ i_c(t) = -132\times10^{-6} \times 9.7 \times 330 \sin(330t) ] [ i_c(t) = -0.4225 \sin(330t)\ A = -422.5 \sin(330t)\ mA ] ✅ **Answer:** [ \boxed{i_c(t) = -422.5\sin(330t)\ \text{mA}} ] --- ### **(b)** Maximum stored energy: [ w_{max} = \frac{1}{2} C (v_{max})^2 ] where (v_{max} = 16.9 + 9.7 = 26.6\ V) [ w_{max} = \frac{1}{2}(132\times10^{-6})(26.6)^2 = 0.0467\ J = 46{,}699\ \mu J ] ✅ **Answer:** [ \boxed{w_{max} = 46{,}699\ \mu J} ] --- ### **(c)** If [ i_c(t) = 19.4 e^{-165t}\ mA,\ v_c(0)=0 ] then [ i_c(t) = C \frac{dv_c(t)}{dt} \implies dv_c(t) = \frac{i_c(t)}{C}dt ] Integrating: [ v_c(t) = \frac{19.4\times10^{-3}}{132\times10^{-6}}\frac{1}{165}(1 - e^{-165t}) ] [ v_c(t) = 0.8907(1 - e^{-165t})\ V = 890.7(1 - e^{-165t})\ mV ] ✅ **Answer:** [ \boxed{v_c(t) = 890.7(1 - e^{-165t})\ \text{mV}} ] --- ### **(d)** If same (i_c(t) = 19.4e^{-165t}\ mA) but (v_c(0)=7.2\ mV), then [ v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ mV ] ✅ **Answer:** [ \boxed{v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ \text{mV}} ] --- ### **(e)** If stored energy (w(t) = 17e^{-560t}\ \mu J) Convert to Joules: (17\times10^{-6} e^{-560t}) [ w(t) = \frac{1}{2} C v_c^2(t) \Rightarrow v_c(t) = \sqrt{\frac{2w(t)}{C}} ] [ v_c(t) = \sqrt{\frac{2(17\times10^{-6} e^{-560t})}{132\times10^{-6}}} = 0.5075 e^{-280t}\ V ] Now current: [ i_c(t) = C \frac{dv_c}{dt} = C(-0.5075)(280)e^{-280t} ] [ i_c(t) = -132\times10^{-6} \times 0.5075 \times 280 e^{-280t} = -18.76\times10^{-3} e^{-280t}\ A ] [ i_c(t) = -18.76 e^{-280t}\ mA ] ✅ **Answers:** [ \boxed{i_c(t) = -18.76 e^{-280t}\ \text{mA}} ] [ \boxed{v_c(t) = 0.5075 e^{-280t}\ \text{V}} ] --- ### ✅ **Final Summary** | 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)=0.5075e^{-280t}) | mA, V |

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Asked by: Jack
Created at: 2025-10-10 09:11:13

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# Capacitor Problem: Step-by-Step Final Answers Given: - Capacitance: \( C = 132\ \mu F = 132 \times 10^{-6}\ \text{F} \) --- ## (a) Find \( i_c(t) \) from \( v_c(t) = 7.2 + 19.4 \cos^2(165t)\ V \) ### Steps 1. Use the 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 for current: \[ i_c(t) = C \frac{dv_c}{dt} = C \left[ -9.7 \cdot 330 \sin(330t) \right] \] \[ i_c(t) = -132 \times 10^{-6} \times 9.7 \times 330 \sin(330t) \] \[ i_c(t) = -.4225 \sin(330t)\ \text{A} = -422.5 \sin(330t)\ \text{mA} \] **Final Answer:** \[ \boxed{ i_c(t) = -422.5\sin(330t)\ \text{mA} } \] --- ## (b) Maximum Stored Energy - \( v_{max} = 16.9 + 9.7 = 26.6\ \text{V} \) \[ w_{max} = \frac{1}{2} C (v_{max})^2 = \frac{1}{2}(132 \times 10^{-6})(26.6)^2 = .0467\ \text{J} = 46,699\ \mu\text{J} \] **Final Answer:** \[ \boxed{ w_{max} = 46,699\ \mu\text{J} } \] --- ## (c) Find \( v_c(t) \) from \( i_c(t) = 19.4 e^{-165t}\ \text{mA} \), \( v_c() = \) ### Steps - Integrate: \[ v_c(t) = \int \frac{i_c(t)}{C} dt = \frac{19.4 \times 10^{-3}}{132 \times 10^{-6}} \int e^{-165t} dt \] \[ = \frac{19.4 \times 10^{-3}}{132 \times 10^{-6} \cdot 165} (1 - e^{-165t}) \] \[ = .8907(1 - e^{-165t})\ \text{V} = 890.7(1 - e^{-165t})\ \text{mV} \] **Final Answer:** \[ \boxed{ v_c(t) = 890.7(1 - e^{-165t})\ \text{mV} } \] --- ## (d) \( v_c() = 7.2\ \text{mV} \), same \( i_c(t) \) as (c) \[ v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ \text{mV} \] **Final Answer:** \[ \boxed{ v_c(t) = 7.2 + 890.7(1 - e^{-165t})\ \text{mV} } \] --- ## (e) Given \( w(t) = 17 e^{-560t}\ \mu\text{J} \), find \( v_c(t) \), \( i_c(t) \) ### Steps 1. Convert to Joules: \( 17 \times 10^{-6} e^{-560t} \) 2. Use \( w(t) = \frac{1}{2} C v_c^2(t) \): \[ v_c(t) = \sqrt{ \frac{2 w(t)}{C} } = \sqrt{ \frac{2 \cdot 17 \times 10^{-6} e^{-560t}}{132 \times 10^{-6}} } \] \[ = .5075 e^{-280t}\ \text{V} \] 3. Differentiate for current: \[ i_c(t) = C \frac{dv_c}{dt} = C \cdot (-.5075 \cdot 280) e^{-280t} \] \[ = -132 \times 10^{-6} \times .5075 \times 280\, e^{-280t} = -18.76 \times 10^{-3} e^{-280t}\ \text{A} \] \[ = -18.76 e^{-280t}\ \text{mA} \] **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) | \( i_c(t) = -18.76 e^{-280t} \), \( v_c(t) = .5075 e^{-280t} \) | mA, V | --- **Note:** All results are rounded to 4 significant digits. Images are not included as all work is shown in LaTeX math.

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