The diode used in the circuit shown in the fi | vedclass

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(B) The maximum current $I$ that the diode can handle is determined by its maximum power rating $P$ and its constant voltage drop $V_d$.Given $P = 100

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$5$

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(B) The maximum current $I$ that the diode can handle is determined by its maximum power rating $P$ and its constant voltage drop $V_d$.Given $P = 100\; mW = 100 	imes 10^{-3}\; W$ and $V_d = 0.5\; V$.The maximum current is $I = \frac{P}{V_d} = \frac{100 	imes 10^{-3}}{0.5} = 0.2\; A$.Applying Kirchhoff's Voltage Law to the circuit,the voltage across the resistor $R$ is $V_R = V_{source} - V_d = 1.5\; V - 0.5\; V = 1.0\; V$.Using Ohm's law,$V_R = I 	imes R$,so $R = \frac{V_R}{I} = \frac{1.0}{0.2} = 5\; \Omega$.

The diode used in the circuit shown in the figure has a constant voltage drop of $0.5\; V$ at all currents and a maximum power rating of $100\; mW$. What should be the value of the resistor $R$,connected in series with the diode,to obtain the maximum current? (in $\Omega$)

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