Question:
The forward resistance for silicon diode carrying a current of \(20~mA\) is
The forward resistance for silicon diode carrying a current of \(20~mA\) is
Show Hint
For silicon diodes,
\[
V_F \approx 0.7V
\]
and for germanium diodes,
\[
V_F \approx 0.3V.
\]
These values are frequently used in numerical problems.
Updated On: Jun 25, 2026
- \(15~\Omega\)
- \(35~\Omega\)
- \(50~\Omega\)
- \(20~k\Omega\)
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The Correct Option is B
Solution and Explanation
Concept:
The DC forward resistance of a diode is defined as
\[
R_f=\frac{V_F}{I_F}
\]
where
\[
V_F=\text{forward voltage}
\]
and
\[
I_F=\text{forward current}.
\]
For a silicon diode,
\[
V_F \approx 0.7V.
\]
Step 1: Write the given values.
\[ V_F=0.7V \] \[ I_F=20mA=0.02A \]
Step 2: Apply the forward resistance formula.
\[ R_f = \frac{V_F}{I_F} \] \[ = \frac{0.7}{0.02} \] \[ = 35\Omega \]
Step 3: State the result.
Therefore, \[ \boxed{R_f=35\Omega} \]
Step 1: Write the given values.
\[ V_F=0.7V \] \[ I_F=20mA=0.02A \]
Step 2: Apply the forward resistance formula.
\[ R_f = \frac{V_F}{I_F} \] \[ = \frac{0.7}{0.02} \] \[ = 35\Omega \]
Step 3: State the result.
Therefore, \[ \boxed{R_f=35\Omega} \]
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