Figure given below shows the waveform of the current passing through an inductor of resistance 1 Ω and inductance 2 H. The energy absorbed by the inductor in the first four seconds is

(a) 144 J (b) 98 J

(c) 132 J (d) 168 J

Step-by-Step

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Power is the rate of change of energy.

Therefore,

E=\int_0^t P d t

We know that

\begin{gathered} P=V I=I \cdot\left\lgroup L \frac{d I}{d t}\right\rgroup\left\lgroup \text { as } V_{ L }=L \frac{d I}{d t}\right\rgroup \\ E=\int_0^t L I\left\lgroup \frac{d I}{d t}\right\rgroup \cdot d t \end{gathered}

Given that \left\{\begin{array}{rlrl} d I / d t & =3 & 0 \leq t<2 \\ & =0 & 2<t<4 \end{array}\right.

\begin{aligned} E & =\int_0^4 2 I \frac{d I}{d t} \cdot d t=2 \int_0^2 3 I \cdot d t+2 \int_0^4 0 \cdot I \cdot d t=6 \cdot \int_0^2 I \cdot d t \\ & =6 \times \text { area under the current curve } \\ & =6 \times \frac{1}{2} \times 2 \times 6=36 J \end{aligned}

Given that \left\{\begin{aligned} I & =3 t & & 0 \leq t \leq 2 \\ & =6 & & 2<t<4 \end{aligned}\right.

For resistor,

\begin{aligned} E & =\int_0^t I^2 \cdot R d t \quad \quad\left[\text { as } P=V I=I \cdot I R=I^2 R\right] \\ & =\int_0^2(3 t)^2 \cdot d t+\int_2^4 6^2 \cdot d t=9 \cdot\left[\frac{t^3}{3}\right]_0^2+36[t]_2^4 \\ & =9 \times \frac{8}{3}+36 \times 2=24+72=96 J \end{aligned}

Therefore, total energy = 36 + 96 = 132 J.

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