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

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(B) The maximum power rating of the diode is $P = 100 \, mW = 0.1 \, W$.The voltage drop across the diode is $V_d = 0.5 \, V$.The maximum current $I$

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$5 \, \Omega$

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(B) The maximum power rating of the diode is $P = 100 \, mW = 0.1 \, W$.The voltage drop across the diode is $V_d = 0.5 \, V$.The maximum current $I$ that can flow through the diode is given by $I = \frac{P}{V_d} = \frac{0.1 \, W}{0.5 \, V} = 0.2 \, A$.According to Kirchhoff's voltage law for the circuit,the total voltage of the battery $V_b = 1.5 \, V$ is equal to the sum of the voltage drop across the resistor $V_R$ and the voltage drop across the diode $V_d$.$V_b = V_R + V_d$$1.5 \, V = I \cdot R + 0.5 \, V$$1.5 \, V = (0.2 \, A) \cdot R + 0.5 \, V$$1.0 \, V = 0.2 \cdot R$$R = \frac{1.0}{0.2} \, \Omega = 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,for obtaining maximum current?

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