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

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(B) Given: Voltage drop across the diode $(V_D) = 0.5 \, V$. Maximum power rating of the diode $(P) = 100 \, mW = 100 	imes 10^{-3} \, W$. Source vol

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$5 \, \Omega$

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(B) Given: Voltage drop across the diode $(V_D) = 0.5 \, V$. Maximum power rating of the diode $(P) = 100 \, mW = 100 	imes 10^{-3} \, W$. Source voltage $(V_S) = 1.5 \, V$.To find the maximum current $(I_{max})$ that the diode can handle:$P = V_D 	imes I_{max}$$I_{max} = \frac{P}{V_D} = \frac{100 	imes 10^{-3} \, W}{0.5 \, V} = 0.2 \, A$.Applying Kirchhoff's Voltage Law $(KVL)$ to the circuit:$V_S = I_{max} 	imes R + V_D$$1.5 \, V = (0.2 \, A) 	imes R + 0.5 \, V$$1.5 - 0.5 = 0.2 	imes R$$1.0 = 0.2 	imes R$$R = \frac{1.0}{0.2} = 5 \, \Omega$.Therefore,the value of the series resistor $R$ is $5 \, \Omega$.

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?

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