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  • The voltage drop across a forward biased diode is 0.7b V. In the following circuit, the voltages across the <img alt="\mathrm{10\Omega }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmat

The voltage drop across a forward biased diode is 0.7b V. In the following circuit, the voltages across the \mathrm{10\Omega } resistance in series with the diode and \mathrm{20\Omega } resistance are:

Option: 1

0.70 V, 4.28 V


Option: 2

3.58 V, 4.28 V 


Option: 3

5.35 V, 2.14 V 


Option: 4

3.58 V, 9.3 V


Answers (1)

best_answer

Let the currents through the \mathrm{20\Omega } (parallel) and \mathrm{10\Omega } (in series with the diode) be \mathrm{I_{1} } and \mathrm{I_{2} } respectively.

Applying Kirchhoff’s second law for closed loop ABEFA, we get

\mathrm{20 \mathrm{I}_1+10\left(\mathrm{I}_1+\mathrm{I}_2\right)-10=0} -------------(i)

Applying Kirchhoff’s second law for closed loop BCDEB, we get

\mathrm{0.7+10 \mathrm{I}_2-20 \mathrm{I}_1=0} ---------------------(ii)

Solving (i) and (ii), we get

\mathrm{\mathrm{I}_1=0.214 \mathrm{~A} \text { and } \mathrm{I}_2=0.358 \mathrm{~A}}

Thus, voltage across the \mathrm{10\Omega } resistance in series with the diode \mathrm{=0.358 \mathrm{~A} \times 10 \Omega=3.58 \mathrm{~V} }

And voltage across the \mathrm{20 \Omega } \mathrm{\text { resistance }=0.214 \mathrm{~A} \times 20 \Omega=4.28 \mathrm{~V}}

Posted by

vishal kumar

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