Question:
In the circuit shown, diode has \(20\Omega\) forward resistance. When \(V_i\) increases from \(8V\) to \(12V\), change in current is \(x\,mA\). Find \(x\).
In the circuit shown, diode has \(20\Omega\) forward resistance. When \(V_i\) increases from \(8V\) to \(12V\), change in current is \(x\,mA\). Find \(x\).
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Always evaluate current separately in OFF and ON regions of diode.
Updated On: Apr 17, 2026
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Correct Answer: 9.09
Solution and Explanation
Concept:
•Diode conducts only when \(V_i>10V\) (cut-in voltage)
•When OFF → current = 0
•When ON → use Ohm’s law with total resistance
Step 1:Total resistance (ON state) \[ R = 200 + 20 = 220\,\Omega \]
Step 2:At \(V_i = 8V\) \[ V_i<10V \Rightarrow \text{diode OFF} \Rightarrow I_1 = 0 \]
Step 3:At \(V_i = 12V\) \[ V_{net} = 12 - 10 = 2V \] \[ I_2 = \frac{2}{220} = 0.00909\,A = 9.09\,mA \]
Step 4:Change in current \[ \Delta I = I_2 - I_1 = 9.09 - 0 = 9.09\,mA \] Conclusion \[ {x = 9.09\,mA} \]
•Diode conducts only when \(V_i>10V\) (cut-in voltage)
•When OFF → current = 0
•When ON → use Ohm’s law with total resistance
Step 1:Total resistance (ON state) \[ R = 200 + 20 = 220\,\Omega \]
Step 2:At \(V_i = 8V\) \[ V_i<10V \Rightarrow \text{diode OFF} \Rightarrow I_1 = 0 \]
Step 3:At \(V_i = 12V\) \[ V_{net} = 12 - 10 = 2V \] \[ I_2 = \frac{2}{220} = 0.00909\,A = 9.09\,mA \]
Step 4:Change in current \[ \Delta I = I_2 - I_1 = 9.09 - 0 = 9.09\,mA \] Conclusion \[ {x = 9.09\,mA} \]
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