Answer :
Final answer:
The potential difference across the 3 μF and 6 μF capacitors, when connected in series to a 12 V battery, is 8 V and 4 V, respectively.
Explanation:
To find the potential difference (P.d.) across two capacitors in series, we need to recognize that the charge (Q) across both capacitors remains the same due to the continuity of their connection. The formula for capacitors in series helps us calculate the total capacitance (Ct) and subsequently the charge (Q) since Q = CtV, where V is the voltage of the battery. In this scenario, the capacitors of 3 μF and 6 μF are connected in series to a 12 V battery.
First, calculate the total capacitance using the formula for capacitors in series: 1/Ct = 1/C1 + 1/C2. Plugging in the values, we get 1/Ct = 1/3 + 1/6 = 1/2, so Ct = 2 μF. The charge across each capacitor, hence, is Q = Ct*V = 2 μF * 12 V = 24 μC.
Since the charge is constant across both capacitors, the voltage across each capacitor can be calculated using V = Q/C. Thus, the P.d. across the 3 μF capacitor is V = 24 μC / 3 μF = 8 V, and the P.d. across the 6 μF capacitor is V = 24 μC / 6 μF = 4 V.
