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  • The current through a device that obeys the relation , is increased linearly from zero to a maximum value <

The current through a device that obeys the relation V=\lambda _{0}l^{2} , is increased linearly from zero to a maximum value I_{0} in a time t_{0}. The average power dissipated in this device is

Option: 1

\frac{1}{4} \lambda_0 \mathrm{I}_0^3


Option: 2

\frac{1}{3} \lambda_0 I_0^3


Option: 3

\frac{1}{2} \lambda_0 \mathrm{I}_0^3


Option: 4

None of the above


Answers (1)

best_answer

The current through the device is given by the current voltage relation

V=\lambda_0 I^2

The power dissipated is

VI=\lambda_0 I^3

The average power is

\begin{aligned} \mathrm{P}_{\text {avg }} & =\frac{\lambda}{\mathrm{t}_0} \int_0^{\mathrm{t}_0} \mathrm{I}^3 \mathrm{dt} \\ & =\frac{\lambda_0}{\mathrm{kt}} \int_0^{\mathrm{I}_0} \mathrm{I}^3 \mathrm{dt} \quad ; \mathrm{I}=\mathrm{kt} \text { and } \mathrm{I}_0=\mathrm{k}^2 \mathrm{t}_0 \\ & =\frac{\lambda_0}{\mathrm{I}_0} \times \frac{\mathrm{I}_0^4}{4} \\ & =\frac{\lambda_0 \mathrm{I}_0^3}{4} \\ & =\frac{1}{4}\left(V_0 \times I_0\right) \end{aligned}

\text { where } V_0=\text { maximum p.d. }=\lambda_0 I_0^3 \text {. }

 

Posted by

Kuldeep Maurya

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