A circuit contains a resistor in series with a capacitor, the series combination being connected across the terminals of a battery. The time constant for charging the capacitor is 2.6 s when the resistor has a resistance of 1.9 x 104 Ω. What would the time constant be if the resistance had a value of 6.4 x 104 Ω?

Question

A circuit contains a resistor in series with a capacitor, the series combination being connected across the terminals of a battery. The time constant for charging the capacitor is 2.6 s when the resistor has a resistance of 1.9 x 104 Ω. What would the time constant be if the resistance had a value of 6.4 x 104 Ω?

Answer 1

time constant=RC

so time constant is directly proportional to resistance.
timeconstant= (6.4/1.9)*2.6 seconds

Answer 2

The time constant (τ) can be determined using the formula τ = RC, where R is the resistance and C is the capacitance.

Given that the resistance (R) is 1.9 x 10^4 Ω and the time constant (τ) is 2.6 s, we can rearrange the formula to find the capacitance (C).

τ = RC

2.6 = (1.9 x 10^4)C

C = 2.6 / (1.9 x 10^4)

C ≈ 1.368 x 10^-4 F

Now, we need to calculate the new time constant (τ) with the resistance value of 6.4 x 10^4 Ω.

τ = RC

τ = (6.4 x 10^4)(1.368 x 10^-4)

τ ≈ 8.7552 s

Therefore, the time constant for charging the capacitor would be approximately 8.7552 s when the resistance has a value of 6.4 x 10^4 Ω.

Answer 3

To find the time constant for charging the capacitor with a different resistance value, we can use the formula for the time constant of an RC circuit:

τ = R * C

Where:
τ = Time constant (in seconds)
R = Resistance (in ohms)
C = Capacitance (in farads)

In the given problem, we are given the time constant (τ) and the resistance (R) values for the initial circuit. Let's substitute the values into the formula:

τ1 = 2.6 s
R1 = 1.9 x 10^4 Ω

Therefore, τ1 = R1 * C
2.6 s = (1.9 x 10^4 Ω) * C

To find the capacitance (C), we can rearrange the equation:

C = τ1 / R1

Next, we need to find the time constant (τ2) for the new resistance value (R2):

R2 = 6.4 x 10^4 Ω

To find the new capacitance value (C2), we can use the same equation:

C2 = τ2 / R2

Since the capacitance (C) is constant in this case, we can set the two equations equal to each other:

C1 = C2

Therefore,

τ1 / R1 = τ2 / R2

Now we can solve for τ2:

τ2 = (τ1 * R2) / R1

Substituting the given values:

τ2 = (2.6 s * (6.4 x 10^4 Ω)) / (1.9 x 10^4 Ω)

Calculating this equation will give us the new time constant (τ2) for the circuit with the higher resistance.