Question

A capacitor that is initially uncharged is connected in series with a resistor and an emf source with E = 120 V and negligible internal resistance. Just after the circuit is completed, the current through the resistor is 6.7×10−5 A . The time constant for the circuit is 4.4 s .'What are the resistance of the resistor and the capacitance of the capacitor?

Answer 1

Resistance = 1.791 x 10^(6) Ω

Capacitance = 2.46 x 10^(-6) F

Step-by-step explanation:

We are given;

EMF; E = 120V

Current; I = 6.7 × 10^(−5) A

Time constant; τ = 4.4 s

Now, just after the circuit is completed, the capacitor acts like a wire and thus we use the loop rule;

So,

E - IR = 0

Let's make the resistance R the subject.

R = E/I = 120/(6.7 × 10^(−5)) = 1.791 x 10^(6) Ω

Now formula for time constant is given as;

τ = RC

Where C is capacitance.

Thus, C = τ/R = 4.4/(1.791 x 10^(6))

C = 2.46 x 10^(-6) F

Answer 2

R = 1.79*10^6 Ω

C = 2.46*10^-6 F

Step-by-step explanation:

Given

emf of the source, ε = 120 V

Current passing through the resistor, I = 6.7*10^-5 A

Time constant for the circuit, τ = 4.4 s

From the information above, we can say that RC = 4.4 s

Also, on applying the loop rule, we get

ε - IR = 0

ε = IR

R = ε / I

R = 120 / 6.7*10^-5

R = 1.79*10^6 Ω

Using the first equation, we can thus solve for C

RC = 4.4 s

C = 4.4 / R

C = 4.4 / 1.79*10^6

C = 2.46*10^-6 F

C = 2.46 μF

Therefore, the resistance and capacitance of the capacitor is respectively, 1.79 MΩ and 2.46 μF