Two capacitors are connected in series and the combination is charged to \(100 \mathrm{V}\). If the voltage across each capacitor is \(50 \mathrm{V}\), how do their capacitances compare?

Short Answer

Expert verified
The capacitance of the two capacitors is the same.

Step by step solution

01

Understand the Concept of Capacitance in a Series

In a series, the total capacitance, \(C_t\), can be given by the equation \(1/C_t = 1/C_1 + 1/C_2\). However in this problem, we are told that the voltage across each capacitor is the same, implying that the capacitances for the two capacitors are the same (i.e., \(C_1 = C_2\)).
02

Rewrite the Equation For Two Equal Capacitances

With that known, we can simplify the previous equation to \(1/C_t = 2/C_1\). From this equation, we can infer that for the series, the total capacitance is half the capacitance of one of the identical capacitors.
03

Analyze in terms of Capacitance and Voltage

The problem statement provided, states that each capacitor in the series has a voltage of 50V. As both capacitors are identical in their capacitance and both hold the same voltage, this implies that if one capacitor were to be removed, the voltage of the remaining capacitor would be equivalent to the combined voltage of the two capacitors, which is 100V. This change in voltage, without modifying the capacitance, would obey Ohm's law.
04

Conclusion

In conclusion, the two capacitors have the same value of capacitance. This is because they are in series, have the same voltage drop, and according to Ohm's law, if the voltage across each capacitor is the same, the capacitances must be equivalent. The charge is distributed in such a way that the voltage across each capacitor is equal as they have equal capacitances.

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