Question 11 Mark
The diode used in the circuit shown in the figure has a constant voltage drop at 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with diode for obtaining maximum current?
Answer
View full question & answer→(c) $5 \Omega$
Explanation: Voltage drop across diode, $V _{ d }=0.5 V$
Power rate of diode, $P _{ d }=100 mW=0.1 W$
Resistance of the diode,
$I _{ d }=\frac{V_d}{R_d}=\frac{0.5}{2.5}=0.2 A$
Maximum current through the diode,
$I _{ d }=\frac{V_d}{R_d}=\frac{0.5}{2.5}=0.2 A$
Applied voltage, V = 1.5 V
$R ^{\prime}=\frac{V}{I_d}=\frac{1.5}{0.2}=7.5 \Omega$
Value of the series resistor, $R = R ^{\prime}- R _{ d }=7.5-2.5=5 \Omega$Required total resistance of the circuit,
Explanation: Voltage drop across diode, $V _{ d }=0.5 V$
Power rate of diode, $P _{ d }=100 mW=0.1 W$
Resistance of the diode,
$I _{ d }=\frac{V_d}{R_d}=\frac{0.5}{2.5}=0.2 A$
Maximum current through the diode,
$I _{ d }=\frac{V_d}{R_d}=\frac{0.5}{2.5}=0.2 A$
Applied voltage, V = 1.5 V
$R ^{\prime}=\frac{V}{I_d}=\frac{1.5}{0.2}=7.5 \Omega$
Value of the series resistor, $R = R ^{\prime}- R _{ d }=7.5-2.5=5 \Omega$Required total resistance of the circuit,
