Mutual Induction — Question 9

Section 30-11 · Physics pages 4968–4969

Question 9 Coupled RL circuits decay through two different magnetic modes Two identical fixed coils each have inductance L and series resistance R, with 0 < M < L. There are no sources. Positive currents enter dotted terminals. The equations are L˙i1 + M˙i2 + Ri1 = 0 and M˙i1 + L˙i2 + Ri2 = 0. L1 L2 i1 i2 positive currents enter dotted ends; M > 0 Schematic; labels, not drawing lengths, specify the data. (a) Find decay times for equal and opposite current modes. (b) For initial currents i1=I0 and i2=0, derive both currents at later times.
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Question 9 - Solution Coupled RL circuits decay through two different magnetic modes L1 L2 i1 i2 positive currents enter dotted ends; M > 0 (a) Method and calculation i+ = i1 + i2, i−= i1 −i2 : (L ± M)˙i± + Ri± = 0. τ+ = (L + M)/R, τ−= (L −M)/R . (b) Interpretation and further calculation i+(0) = i−(0) = I0 ⇒i± = I0e−t/τ±. i1 = I0 2 (e−t/τ+ + e−t/τ−), i2 = I0 2 (e−t/τ+ −e−t/τ−) . Physical check The secondary initially rises in the positive direction as it opposes the falling primary flux; both currents ultimately dissipate away.

Question 9

Coupled RL circuits decay through two different magnetic modes

Two identical fixed coils each have inductance L and series resistance R, with 0<M<L0<M<L. There are no sources. Positive currents enter dotted terminals. The equations are Li̇1+Mi̇2+Ri1=0L\dot i_1+M\dot i_2+Ri_1=0 and Mi̇1+Li̇2+Ri2=0M\dot i_1+L\dot i_2+Ri_2=0.

Diagram shown in the original SVG worksheet.

  1. Find decay times for equal and opposite current modes.

  2. For initial currents i1=I0 and i2=0, derive both currents at later times.

Question 9 - Solution

Coupled RL circuits decay through two different magnetic modes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

i+=i1+i2,i−=i1−i2:(L±M)i̇±+Ri±=0.i_+=i_1+i_2,\quad i_-=i_1-i_2:\quad(L\pm M)\dot i_\pm+Ri_\pm=0.τ+=(L+M)/R,τ−=(L−M)/R.\boxed{\tau_+=(L+M)/R,\quad\tau_-=(L-M)/R}.

(b) Interpretation and further calculation

i+(0)=i−(0)=I0⇒i±=I0e−t/τ±.i_+(0)=i_-(0)=I_0\Rightarrow i_\pm=I_0e^{-t/\tau_\pm}.i1=I02(e−t/τ++e−t/τ−),i2=I02(e−t/τ+−e−t/τ−).\boxed{i_1=\tfrac{I_0}{2}(e^{-t/\tau_+}+e^{-t/\tau_-}),\quad i_2=\tfrac{I_0}{2}(e^{-t/\tau_+}-e^{-t/\tau_-})}.

Physical check

The secondary initially rises in the positive direction as it opposes the falling primary flux; both currents ultimately dissipate away.