Question

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

• A. 0.70 V, 4.28 V
• B. 3.58 V, 4.28 V
• C. 5.35 V, 2.14 V
• D. 3.58 V, 9.3 V

Answer 1

Answer: (B)

Let the currents through the $20\Omega$ (parallel) and $20\Omega$ (in series with the diode) be $I_1$ and $I_2$ respectively.

Applying Kirchhoff’s second law for closed loop ABEFA, we get
$20I_{1} + 10\left(I_{1} + I_{2}\right) - 10 = 0\quad.... \left(i\right)$
Applying Kirchhoff’s second law for closed loop BCDEB, we get
$0.7 + 10I_{2} - 20I_{1} = 0\quad.... \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get
$I_{1} = 0.214 \,A$ and $I_{2} = 0.358\, A$
Thus, voltage across the $10\,\Omega$ resistance in series with the diode $= 0.358\, A × 10\,\Omega = 3.58 \,V$
And voltage across the $20\,\Omega$ resistance
$= 0.214\, A × 20\,\Omega = 4.28 \,V$