The total heat produced in resistor R in an R | vedclass

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(B) The power dissipated in the resistor is given by $P = I^2 R$.Since the current through the resistor varies with time,we must integrate the power o

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$\frac{1}{2}LI_0^2$

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(B) The power dissipated in the resistor is given by $P = I^2 R$.Since the current through the resistor varies with time,we must integrate the power over time to find the total heat energy $W$.$W = \int_{0}^{\infty} I^2 R \, dt$The current in an $RL$ circuit during decay is given by $I = I_0 e^{-(R/L)t}$.Substituting this into the integral:$W = \int_{0}^{\infty} (I_0 e^{-(R/L)t})^2 R \, dt = \int_{0}^{\infty} I_0^2 R e^{-(2R/L)t} \, dt$$W = I_0^2 R \left[ \frac{e^{-(2R/L)t}}{-(2R/L)} \right]_{0}^{\infty} = I_0^2 R \left( 0 - \left( -\frac{L}{2R} \right) \right) = \frac{1}{2} L I_0^2$.Alternatively,by the law of conservation of energy,the total heat produced in the resistor must equal the total energy initially stored in the magnetic field of the inductor,which is $U = \frac{1}{2} L I_0^2$.

The total heat produced in resistor $R$ in an $RL$ circuit when the current in the inductor decreases from $I_0$ to $0$ is

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