Điện áp trên tụ điện 5 mF khi có dòng điện i(t) = 7sin(100t + 300) A chạy qua là:

The voltage across a capacitor is related to the current by the formula i(t) = C * dv(t)/dt. Therefore, v(t) = (1/C) * integral(i(t) dt). In this case, i(t) = 7sin(100t + 300) A and C = 5 mF = 5 * 10^(-3) F. Thus, v(t) = (1/(5 * 10^(-3))) * integral(7sin(100t + 300) dt) = 200 * integral(7sin(100t + 300) dt) = 1400 * integral(sin(100t + 300) dt). The integral of sin(100t + 300) is -cos(100t + 300)/100. Therefore, v(t) = 1400 * (-cos(100t + 300)/100) = -14cos(100t + 300) V. Using the identity -cos(x) = sin(x - 90°), we have: v(t) = 14sin(100t + 300 - 90°) = 14sin(100t + 210°) V. However, none of the options perfectly match. Since -cos(x) = cos(x + 180), v(t) = 14cos(100t + 300 + 180) = 14cos(100t + 540). Also, sin(100t + 210) = -sin(100t + 210 + 180) = -sin(100t + 390). There seems to be an error in the options provided as none match this result.