A 12μF capacitor and a 4μF capacitor are connected in series. The ratio of voltages across them will be:
1:3
To find the ratio of voltages across two capacitors connected in series, we need to use the concept that the charge on each capacitor in series is the same. The voltage across a capacitor \(V\) is given by the formula \(V = \frac{Q}{C}\), where \(Q\) is the charge and \(C\) is the capacitance.
Let's denote:
- Capacitance of the first capacitor, \(C_1 = 12 \, \mu\text{F}\)
- Capacitance of the second capacitor, \(C_2 = 4 \, \mu\text{F}\)
Since the capacitors are in series, the charge \(Q\) on each capacitor is the same. So we can write: \(V_1 = \frac{Q}{C_1}\) and \(V_2 = \frac{Q}{C_2}\)
The ratio of the voltages across these capacitors is: \(\frac{V_1}{V_2} = \frac{\frac{Q}{C_1}}{\frac{Q}{C_2}} = \frac{C_2}{C_1}\)
Substituting the capacitance values: \(\frac{V_1}{V_2} = \frac{4 \, \mu\text{F}}{12 \, \mu\text{F}} = \frac{1}{3}\)
Therefore, the ratio of voltages across the 12μF capacitor and the 4μF capacitor is 1:3.
Thus, the correct answer is 1:3.
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