Exam Question:

A circuit consists of a battery with an electromotive force (emf) of 12 V and internal resistance of 4 Ω, connected in series with a resistor, capacitor, and inductor, as shown below:

The resistor has a resistance of 6 Ω, the capacitor has a capacitance of 10 μF, and the inductor has an inductance of 2 H. The switch is closed at time t = 0, and a steady current is reached after a long time. Determine the following:

a) The current in the circuit when the switch is first closed. b) The time constant of the circuit. c) The value of the current reached when the circuit reaches a steady state.

Answer with Step-by-Step Explanation:

a) To find the current in the circuit when the switch is first closed, we need to calculate the total resistance of the circuit, and then use Ohm's Law (V = IR) to determine the current.

The total resistance (RT) of the circuit is given by:

RT = r + R + XL

where r is the internal resistance of the battery, R is the resistance of the resistor, and XL is the reactance of the inductor.

Given values: r = 4 Ω, R = 6 Ω, XL = ωL

where ω is the angular frequency and L is the inductance.

Using the formula for angular frequency (ω = 2πf), where f is the frequency, and given that the switch is closed at time t = 0 (i.e., at the beginning), we consider the steady state current. In the steady state, the voltage across the inductor is zero, so the reactance XL is zero.

Hence, the total resistance will simply be the sum of the internal resistance and the resistance of the resistor:

RT = r + R

Substituting the given values:

RT = 4 Ω + 6 Ω = 10 Ω

Applying Ohm's Law using the total resistance:

V = IR

12 V = I × 10 Ω

Solving for current (I):

I = 12 V / 10 Ω = 1.2 A

Therefore, the current in the circuit when the switch is first closed is 1.2 A.

b) The time constant of the circuit (τ) can be calculated using the formula:

τ = RC

where R is the resistance and C is the capacitance.

Given values: R = 6 Ω, C = 10 μF = 10 × 10^(-6) F

Substituting the values into the formula:

τ = (6 Ω) × (10 × 10^(-6) F) = 60 × 10^(-6) s = 60 μs

Therefore, the time constant of the circuit is 60 μs.

c) When the circuit reaches a steady state, the voltage across the capacitor is zero, so the reactance XC is zero. At steady state, the inductor behaves like a short circuit and can be ignored. The total resistance is then simply the sum of the internal resistance and the resistance of the resistor:

RT = r + R = 4 Ω + 6 Ω = 10 Ω

Applying Ohm's Law using the total resistance:

V = IR

Substituting the given values:

12 V = I × 10 Ω

Solving for current (I):

I = 12 V / 10 Ω = 1.2 A

Therefore, the value of the current reached when the circuit reaches a steady state is 1.2 A.