Two capacitors if capacitance 0.4\(\mu F\) and 0.5\(\mu F\) are connected in parallel and charged to a p.d. of 50V, Determine the total charge acquired

Answer Details

When two capacitors are connected in parallel, their total capacitance is the sum of their individual capacitances. Therefore, the total capacitance in this case is: C_total = C_1 + C_2 = 0.4\(\mu F\) + 0.5\(\mu F\) = 0.9\(\mu F\) The charge acquired by each capacitor is given by: Q = C * V where Q is the charge, C is the capacitance, and V is the voltage. For the 0.4\(\mu F\) capacitor, the charge acquired is: Q_1 = C_1 * V = 0.4\(\mu F\) * 50V = 20\(\mu C\) For the 0.5\(\mu F\) capacitor, the charge acquired is: Q_2 = C_2 * V = 0.5\(\mu F\) * 50V = 25\(\mu C\) Therefore, the total charge acquired is the sum of the charges acquired by each capacitor: Q_total = Q_1 + Q_2 = 20\(\mu C\) + 25\(\mu C\) = 45\(\mu C\) Therefore, the correct answer is 45\(\mu C\).