The value of the resistor,R_S,needed in the d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In the given $dc$ voltage regulator circuit,the total current flowing through the series resistor $R_S$ is the sum of the load current $I_L$ and t

Vedclass · Accepted Answer

$\frac{(V_i - V_L)}{(n + 1)I_L}$

Vedclass · Answer

(A) In the given $dc$ voltage regulator circuit,the total current flowing through the series resistor $R_S$ is the sum of the load current $I_L$ and the Zener diode current $I_Z$.From the circuit diagram,the Zener diode current is given as $I_Z = nI_L$.Therefore,the total current $I$ flowing through $R_S$ is $I = I_L + I_Z = I_L + nI_L = (n + 1)I_L$.The voltage drop across the resistor $R_S$ is $V_i - V_L$.Using Ohm's law,$V_i - V_L = I 	imes R_S$.Substituting the value of $I$,we get $V_i - V_L = (n + 1)I_L 	imes R_S$.Thus,the value of the resistor $R_S$ is $R_S = \frac{V_i - V_L}{(n + 1)I_L}$.

The value of the resistor,$R_S$,needed in the $dc$ voltage regulator circuit shown here,equals

Similar Questions

$A$ diode has Zener voltage of $10 \text{ V}$ and maximum power dissipation of $0.5 \text{ W}$. The minimum resistance to be used in series with this diode for safety when it is connected to a $25 \text{ V}$ power supply is . . . . . . $\Omega$.

$A$ Zener diode, having a breakdown voltage of $15 \text{ V}$, is used in a voltage regulator circuit as shown. The current through the Zener diode is (in $\text{ mA}$)

The value of power dissipated across the Zener diode $(V_{Z} = 15 \text{ V})$ connected in the circuit as shown in the figure is $x \times 10^{-1} \text{ watt}$. The value of $x$, to the nearest integer, is ..............

$A$ reverse biased Zener diode when operated in the breakdown region works as

In the circuit given,the current through the Zener diode is ....... $mA$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module