Question

Given a series RC circuit set up by R=1M2, Vps=1v and 2 capacitors C, = 1pF, C2 = 1uF. Assuming one

capacitor is being tested in a RC circuit at a time, which capacitor has faster charging speed?

(1 pF = 1 × 10-12F)

Answer 1

In an RC circuit, the charging speed of a capacitor depends on its capacitance and the resistance in the circuit. The time constant, denoted as τ (tau), is the product of the resistance and the capacitance in the circuit.

For the first capacitor with a capacitance of 1pF, and assuming the resistance remains the same at 1M2 (which is equivalent to 1.2MΩ), the time constant would be:

τ = R * C = (1.2MΩ) * (1pF) = 1.2 × 10^6 Ω * 1 × 10^(-12) F = 1.2 × 10^(-6) seconds

For the second capacitor with a capacitance of 1uF, the time constant would be:

τ = R * C = (1.2MΩ) * (1uF) = 1.2 × 10^6 Ω * 1 × 10^(-6) F = 1.2 seconds

Comparing the two time constants, we can see that the first capacitor with 1pF has a much smaller time constant than the second capacitor with 1uF.

Therefore, the first capacitor with 1pF has a faster charging speed than the second capacitor with 1uF.

In conclusion, when only one capacitor is being tested at a time in the RC circuit, the capacitor with a smaller capacitance will have a faster charging speed.