Question

A circuit contains a D-cell battery, a switch, a 20-Ωresistor, and three 20-mF capacitors. The capacitors areconnected in parallel, and the parallel connection ofcapacitors are connected in series with the switch, theresistor and the battery. (a) What is the equivalentcapacitance of the circuit? (b) What is the RC timeconstant? (c) How long before the current decreases to50% of the initial value once the switch is closed?

Answer 1

Step-by-step explanation:

(a) To find the equivalent capacitance of the circuit, we can use the formula for the equivalent capacitance of capacitors in parallel:

C_eq = C1 + C2 + C3

where C1, C2, and C3 are the capacitances of the three capacitors. Substituting the given values, we get:

C_eq = 20 mF + 20 mF + 20 mF = 60 mF = 6.0 × 10^-5 F

Therefore, the equivalent capacitance of the circuit is 6.0 × 10^-5 F.

(b) To find the RC time constant of the circuit, we can use the formula:

τ = RC

where R is the resistance of the resistor and C is the equivalent capacitance of the circuit. Substituting the given values, we get:

τ = (20 Ω)(6.0 × 10^-5 F) = 1.2 × 10^-3 s

Therefore, the RC time constant of the circuit is 1.2 × 10^-3 s.

(c) To find the time it takes for the current to decrease to 50% of the initial value, we can use the formula:

I = I0 e^(-t/τ)

where I0 is the initial current, I is the current after a time t, τ is the RC time constant, and e is the mathematical constant approximately equal to 2.71828. Solving for t, we get:

t = -τ ln(I/I0)

Substituting the given values, we get:

t = -(1.2 × 10^-3 s) ln(0.5) = 8.3 × 10^-4 s

Therefore, it takes 8.3 × 10^-4 s for the current to decrease to 50% of the initial value once the switch is closed.