Diode Off State in Circuit Analysis

0.212 mA

18.6 V

3.32 mA

<p>3.11 mA</p> Signup and view all the answers

<p>Forward biased, behaves like a closed switch</p> Signup and view all the answers

<p>Reverse biased, appears as open circuit</p> Signup and view all the answers

<p>The diode is in the 'off' state when the resulting current direction does not match the arrow in the diode symbol.</p> Signup and view all the answers

<p>The voltage across the resistor is 0 V when the diode current is 0 A.</p> Signup and view all the answers

<p>The voltage across the resistor is equal to the current through it multiplied by its resistance.</p> Signup and view all the answers

<p>Kirchhoff's voltage law states that the sum of the voltages in a closed loop is equal to zero.</p> Signup and view all the answers

<p>A diode is in the 'on' state when the applied voltage establishes a current in the direction matching the arrow of the diode symbol.</p> Signup and view all the answers

<p>The voltage across the open circuit is 8 V.</p> Signup and view all the answers

<p>The current direction matches with the arrow symbol of the diode, and the applied voltage is greater than 0.7 V.</p> Signup and view all the answers

<p>$V_o = 0.7 V$</p> Signup and view all the answers

<p>$I_1 = \frac{E - V_D}{R}$</p> Signup and view all the answers

<p>$I_D1 = I_D2 = \frac{I_1}{2}$</p> Signup and view all the answers

<p>$I_D = 13.97$ mA</p> Signup and view all the answers

<p>Both diodes are turned 'on' as indicated by the current directions in the network.</p> Signup and view all the answers