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The barrier potential in a p-n junction diode is 0.3 volts . The current required is 6 mA. If a resistance of `200 Omega` is connected in series with the junction diode then the e.m.f. of the cell required for use in the circuit isA. 0.3 VB. 1.2 VC. 0.9 VD. 1.5 V |
Answer» Correct Answer - D `V_(b)=0.3 V, I=6 mA , R= 200 Omega , E=?` `I=(E-V_(b))/(R)` `E=IR+V_(b)` `=6xx10^(-3) xx200+0.3` `=1.2+0.3` `=1.5 V` |
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