Let's break down the problem step by step: **Given:** - The voltage across the capacitor: \( V = 2.5 \sin(10t) \) V - Capacitance: \( C = 5 \) F - **Formula for capacitor current:** \( I_C = C \frac{dV}{dt} \) --- ### Step 1: Differentiate the Voltage Given \( V = 2.5 \sin(10t) \): \[ \frac{dV}{dt} = 2.5 \frac{d}{dt}[\sin(10t)] = 2.5 \cdot 10 \cos(10t) = 25 \cos(10t) \] --- ### Step 2: Substitute into the Capacitor Current Formula \[ I_C = C \frac{dV}{dt} = 5 \cdot 25 \cos(10t) = 125 \cos(10t) \] --- ### **Final Answer** \[ \boxed{I_C(t) = 125 \cos(10t) \ \text{A}} \] --- **Notes:** - The derivative identity you mentioned, \(\sin' x = +(1 - \cos^2 x)\), is not standard for the derivative of sine. The standard derivative is \(\frac{d}{dx}\sin x = \cos x\). - The current is in amperes if voltage is in volts and capacitance is in farads. - If you need the **peak current**, it's 125 A. - If you want the **RMS current**, it's \( \frac{125}{\sqrt{2}} \) A. Let me know if you need further explanation!

Let's carefully analyze and solve the problem step by step. --- ### Given Data: - Voltage across the capacitor: \[ V(t) = 2.5 \sin(101 t) \ \text{V} \] - Capacitance: \[ C = 55 \ \text{F} \] - The derivative of sine: \[ \frac{d}{dt} \sin x = \cos x \] - The problem mentions \(\sin' x = + (1 - \cos^2 x)\), which is a misstatement or a hint to use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). But for derivatives, the standard rule applies. --- ### Objective: Find the **capacitor current** \( I_C(t) \), which is given by: \[ I_C(t) = C \frac{dV(t)}{dt} \] --- ### Step 1: Differentiate \( V(t) \) \[ V(t) = 2.5 \sin(101 t) \] Using the chain rule: \[ \frac{dV}{dt} = 2.5 \times 101 \cos(101 t) = 252.5 \cos(101 t) \] --- ### Step 2: Calculate the capacitor current \( I_C(t) \) \[ I_C(t) = C \frac{dV}{dt} = 55 \times 252.5 \cos(101 t) \] \[ I_C(t) = (55 \times 252.5) \cos(101 t) \] Calculate the coefficient: \[ 55 \times 252.5 = 55 \times (250 + 2.5) = 55 \times 250 + 55 \times 2.5 = 13,750 + 137.5 = 13,887.5 \] --- ### **Final Expression:** \[ \boxed{ I(t) = 13,887.5 \cos(101 t) \ \text{A} } \] --- ### **Summary:** - The capacitor current oscillates with the same frequency as the voltage but shifted in phase (cosine vs. sine). - The amplitude of the current is **13,887.5 A**. --- **Note:** The extremely high current value suggests either very large capacitance or high voltage amplitude; in practical systems, such values might be scaled or simulated. Let me know if you'd like further clarification!