A Zener diode voltage regulator operated in the range 120 V -180 V produces a constant supply of 110 V and 250 mA to the load. If the maximum current is equally shared between the load and the Zener diode, then the values of load resistance (RL) and series resistance (RS) are respectively

Question

A Zener diode voltage regulator operated in the range 120 V -180 V produces a constant supply of 110 V and 250 mA to the load. If the maximum current is equally shared between the load and the Zener diode, then the values of load resistance (RL) and series resistance (RS) are respectively (A) RL=280 Ω, RS=70 Ω (B) RL=440 Ω, RS=140 Ω (C) RL=70 Ω, RS=280 Ω (D) RL=440 Ω, RS=1400 Ω

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b6a43309e6b8eef395b5395fbbb0f13e-.png" /> RL=(VL/II .)=(110/250 × 10-3)=440 Ω As current is equally shared by Zener diode and resistance R. Then, I=2 × 250=500 mA =0.5 A So, RS=(VS/I)=((180-110)/0.5) =(70/0.5)=140 Ω

$R_{L} = 280 Ω, R_{S} = 70 Ω$

$R_{L} = 440 Ω, R_{S} = 140 Ω$

$R_{L} = 70 Ω, R_{S} = 280 Ω$

$R_{L} = 440 Ω, R_{S} = 1400 Ω$

$R_{L} = \frac{V _{L}}{I _{I .}} = \frac{110}{250 \times 1 0 ^{- 3}} = 440 Ω$
As current is equally shared by Zener diode and resistance $R$ .
Then, $I = 2 \times 250 = 500 m A$
$= 0.5 A$
So, $R_{S} = \frac{V _{S}}{I} = \frac{( 180 - 110 )}{0.5}$
$= \frac{70}{0.5} = 140 Ω$

Q. A Zener diode voltage regulator operated in the range 120V−180V produces a constant supply of 110V and 250mA to the load. If the maximum current is equally shared between the load and the Zener diode, then the values of load resistance (RL​) and series resistance (RS​) are respectively

RL​=280Ω,RS​=70Ω

RL​=440Ω,RS​=140Ω

RL​=70Ω,RS​=280Ω

RL​=440Ω,RS​=1400Ω

RL​=II.​VL​​=250×10−3110​=440Ω As current is equally shared by Zener diode and resistance R. Then, I=2×250=500mA =0.5A So, RS​=IVS​​=0.5(180−110)​ =0.570​=140Ω

$R_{L} = 280 Ω, R_{S} = 70 Ω$

$R_{L} = 440 Ω, R_{S} = 140 Ω$

$R_{L} = 70 Ω, R_{S} = 280 Ω$

$R_{L} = 440 Ω, R_{S} = 1400 Ω$

$R_{L} = \frac{V _{L}}{I _{I .}} = \frac{110}{250 \times 1 0 ^{- 3}} = 440 Ω$
As current is equally shared by Zener diode and resistance $R$ .
Then, $I = 2 \times 250 = 500 m A$
$= 0.5 A$
So, $R_{S} = \frac{V _{S}}{I} = \frac{( 180 - 110 )}{0.5}$
$= \frac{70}{0.5} = 140 Ω$