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
$20I_1+10\left(I_1+I_2\right)-10=0....\left(i\right)$
Applying Kirchhoff’s second law for closed loop BCDEB, we get
$0.7+10I_2-20I_1=0....\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get
$I_1=0.214A$ and $I_2=0.358A$
Thus, voltage across the $10\Omega$ resistance in series with the diode $=0.358A\times 10\Omega =3.58V$
And voltage across the $20\Omega$ resistance
$=0.214A\times 20\Omega =4.28V$