Get Answers to all your Questions

header-bg qa

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 resistance R connected in series with diode for obtaining maximum current?

Option: 1

\mathrm{2.5 \Omega}


Option: 2

\mathrm{10 \Omega}


Option: 3

\mathrm{12.5 \Omega}


Option: 4

\mathrm{15 \Omega}


Answers (1)

best_answer

Voltage drop across diode, \mathrm{\mathrm{V}_{\mathrm{d}}=0.5 \mathrm{~V}}

Maximum power rating, \mathrm{\begin{aligned} & P=100 \mathrm{~mW} \\ & =100 \times 10^{-3} \mathrm{~W}=0.1 \mathrm{~W} \end{aligned}}

Diode resistance, \mathrm{\mathrm{R}_{\mathrm{d}}=\frac{\mathrm{V}_{\mathrm{d}}^2}{\mathrm{P}}=\frac{(0.5 \mathrm{~V})^2}{0.1 \mathrm{~W}}=2.5 \Omega}

Current in diode, \mathrm{\mathrm{I}_{\mathrm{d}}=\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{R}_{\mathrm{d}}}=\frac{0.5 \mathrm{~V}}{2.5 \Omega}=0.2 \mathrm{~A}}

Total resistance of the circuit \mathrm{=\frac{2.5 \mathrm{~V}}{0.2 \mathrm{~A}}=12.5 \Omega}

Resistance R in circuit, \mathrm{\mathrm{R}=12.5 \mathrm{~W}-2.5 \mathrm{~W}=10 \mathrm{~W}}

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE