A battery is connected in series with a R = 0.26 Ω resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is τ = 0.10 s, and the maximum current in the circuit is I = 8.4 A. Find the following.

(a) the emf of the battery V?
(b) the inductance of the circuit
mH ?
(c) the current in the circuit after one time constant has elapsed
A ?

(d) the voltage across the resistor and the voltage across the inductor after one time constant has elapsed
resistor V
inductor V

a) as t--> oo inductor is a short circuit, no reactance

so
V = i at big t * R
V = (8.4)(.26) = 2.14 Volts

b) at t = 0, i = 0, so all voltage across L
2.14 = L di/dt

let i = 8.4 [ 1 - e^-t/tau ]
given tau = .1 s
i = 8.4 [ 1 - e^-10 t ]
di/dt = 8.4 (10 e^-10 t)
di/dt = 84 at t = 0
so
2.14 = L (84)
L = .0255 H

c) when t = tau = .1
i = 8.4 ( 1 - e^-1) = 8.4(.632) = 5.31 amps

d) Vresistor = i R = 5.31*.26 = 1.38 volts
Vinductor = 2.14 - 1.38 = .759 volts

To find the values requested, we can use the formulas and principles of circuit analysis.

(a) The emf of the battery (V)

In a series circuit, the total voltage across all components is equal to the sum of the voltage drops across each individual component. In this case, the only components are the resistor and the inductor.

Using Ohm's Law, we know that the voltage across a resistor (Vr) is equal to the product of the current (I) and the resistance (R): Vr = I * R.

Since the resistor and inductor are in series, the total voltage across the circuit (V) is equal to the voltage across the resistor (Vr).

Given that the current (I) is 8.4 A and the resistance (R) is 0.26 Ω, we can calculate the voltage (V) as follows:
V = Vr = I * R = 8.4 A * 0.26 Ω = 2.184 V.

Therefore, the emf of the battery is 2.184 V.

(b) The inductance of the circuit (L)

The time constant (τ) is a property of the circuit defined as the product of the resistance (R) and the inductance (L): τ = L / R.

Given that the time constant (τ) is 0.10 s and the resistance (R) is 0.26 Ω, we can rearrange the formula to solve for the inductance (L):
L = τ * R = 0.10 s * 0.26 Ω = 0.026 H.

Therefore, the inductance of the circuit is 0.026 H or 26 mH.

(c) The current in the circuit after one time constant has elapsed (I')

After one time constant (τ), the current in an RL circuit will have reached approximately 63.2% of its maximum value (I).

Using this information, we can calculate the current (I') as follows:
I' = 0.632 * I = 0.632 * 8.4 A = 5.318 A.

Therefore, the current in the circuit after one time constant has elapsed is approximately 5.318 A.

(d) The voltage across the resistor and the voltage across the inductor after one time constant has elapsed

The voltage across a resistor (Vr') can be calculated using Ohm's Law:
Vr' = I' * R = 5.318 A * 0.26 Ω = 1.38108 V.

The voltage across an inductor (Vl') can be calculated using the formula Vl' = L * di/dt, where di/dt is the rate of change of current with respect to time. After one time constant has elapsed, the rate of change of current will have reached its maximum value, equal to I / τ.

Using this information, we can calculate the voltage across the inductor (Vl') as follows:
Vl' = L * (I / τ) = 0.026 H * (8.4 A / 0.10 s) = 2.184 V.

Therefore, the voltage across the resistor after one time constant has elapsed is approximately 1.38108 V, and the voltage across the inductor is approximately 2.184 V.