A diode and a resistor of resistance \(R\) are connected in series across a \(5\) V battery. The cut-in voltage of the diode is \(0.7\) V. If the current through the diode is \(1.0\) mA, what is the value of resistance \(R\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4.3\times10^{3}\,\Omega\)
When the diode conducts, the voltage across it is its cut-in voltage; the rest of the battery voltage appears across the resistor (Kirchhoff’s loop rule).
\( \displaystyle V_R = V - V_D = 5\,\mathrm{V} - 0.7\,\mathrm{V} = 4.3\,\mathrm{V} \)
\( \displaystyle I = 1.0\,\mathrm{mA} = 1.0\times10^{-3}\,\mathrm{A} \)
\( \displaystyle R = \frac{V_R}{I} = \frac{4.3\,\mathrm{V}}{1.0\times10^{-3}\,\mathrm{A}} \)
\( \displaystyle R = 4.3\times10^{3}\,\Omega \)