Answer to AP Physics Free Response Question on Inductance

A free response question involving inductance was posted on 23^rd December 2007 for your practice. This was the question:

In the circuit shown in the figure, the battery and the inductor have negligible resistance. The currents through R₁, R₂ and R₃ are i₁, i₂ and i₃ respectively. The switch S is closed at t = 0.

(a) Calculate the currents i₁, i₂ and i₃ immediately after closing the switch S (when t = 0).

(b) Calculate the currents i₁, i₂ and i₃ when steady state is reached.

(c) Show the nature of variation of the current through the battery graphically, with time ‘t’ on the X-axis and the current (i₁) on the Y-axis.

(d) What are the values of the slopes of the above graph at t = 0 and at t = ∞.

(e) The switch S is opened and the resistance R₃ is disconnected. If the switch S is now closed, what time is required for the current to attain 63.2% of the final steady value?

As promised, I give below the answer:

(a) Since the parallel branch containing R₃ does not contain any inductance, the current through that branch (i₃) rises to the maximum possible value in this condition immediately on closing the switch S. The same current now flows through R₁ since the initial current i₂ ( at t = 0) through the branch containing the inductance L and the resistance R₂ is zero. [Note that in an LR circuit, the current I at the instant ‘t’ is given by I = I₀(1– e^–Rt/L) where I₀ is the final steady current]

Therefore, when t = 0, i₁ = i₃ = (12 V)/ (8 Ω + 8 Ω ) = 0.75 A and i₂ = 0

(b) Steady state is reached when t = ∞. In this condition, the current is controlled by the resistances only [since dI/dt is zero and no opposing induced voltage L(dI/dt) is developed across the inductance]. The circuit then behaves as made of the parallel combination of R₂ and R₃ (having effective value of 4 Ω) in series with the battery and the resistance R₁. The effective resistance of the circuit is (8 + 4) Ω = 12 Ω and the current delivered by the battery (i₁) is (12 V)/ (12 Ω) = 1 A.

Since this current is equally divided between the two parallel branches, current through each branch is 0.5 A. So, i₂ = i₃ = 0.5 A.

(c) The variation of the current i₁ with time t is shown in the adjoining graph. [The current grows exponentially in the inductive branch and this is why the current delivered by the battery finally rises to the steady value of 1 A as shown].

(d) The slope of the graph when t = o is infinity since the current grows abruptly to 0.75 A because of the purely resistive parallel branch containing R₃. The slope when t = ∞ is zero since the current has settled to the final steady value.

(e) When R₃ is disconnected, We have a series LR circuit with L = 2 H and R = R₁ + R₂ = 16 Ω. The time constant of the circuit is L/R = 2/16 = 0.125 s. The current grows to 63.2 % of the final steady current within this time.

[Time constant is L/R which is the time required for the current to become (1 – 1/e) times the final steady current. This is obtained by putting t = L/R in the equation,

I = I₀(1– e^–Rt/L)].

Merry Christmas!