What is the time constant of a circuit with two 220-microfarad capacitors and two 1-megohm resistors in series?

What is the time constant of a circuit with two 220-microfarad capacitors and two 1-megohm resistors in series?

• A. 22 seconds
• B. 110 seconds
• C. 220 seconds
• D. 330 seconds

Answer: (C)

To determine the time constant of the circuit, it's important to understand how to calculate it in an RC (resistor-capacitor) circuit. The time constant, denoted as τ (tau), is given by the formula: τ = R * C where R is the resistance in ohms and C is the capacitance in farads. In this scenario, you have two capacitors with a capacitance of 220 microfarads each. When capacitors are connected in parallel, the total capacitance is the sum of their values, which means: C_total = 220µF + 220µF = 440µF Since this is a microfarad value, converting it to farads gives: C_total = 440 x 10^-6 F. Next, you have two resistors each at 1 megohm. When resistors are in series, their resistances add together: R_total = 1MΩ + 1MΩ = 2MΩ. Converting this to ohms: R_total = 2 x 10^6Ω. Now, substituting these values into the time constant formula: τ = R * C τ = (2 x 10

To determine the time constant of the circuit, it's important to understand how to calculate it in an RC (resistor-capacitor) circuit. The time constant, denoted as τ (tau), is given by the formula:

τ = R * C

where R is the resistance in ohms and C is the capacitance in farads.

In this scenario, you have two capacitors with a capacitance of 220 microfarads each. When capacitors are connected in parallel, the total capacitance is the sum of their values, which means:

C_total = 220µF + 220µF = 440µF

Since this is a microfarad value, converting it to farads gives:

C_total = 440 x 10^-6 F.

Next, you have two resistors each at 1 megohm. When resistors are in series, their resistances add together:

R_total = 1MΩ + 1MΩ = 2MΩ.

Converting this to ohms:

R_total = 2 x 10^6Ω.

Now, substituting these values into the time constant formula:

τ = R * C

τ = (2 x 10