In the given circuit if the power rating of Zener diode is 10 mW, the value of series resistance $R_{s}$ to regulate the input unregulated supply is [2024]

Question

Smart Achievers · Accepted Answer

(1)

Potential difference across $R_{s}$

$V_{1} = 8 - 5 = 3 V$

Current through the load resistor

$I = \frac{5}{1 \times 10^{3}} = 5 mA$

Maximum current through Zener diode

$I_{z max.} = \frac{10}{5} = 2 mA$

and minimum current through Zener diode

$I_{z min.} = 0$

$∴$ $I_{s max.} = 5 + 2 = 7 mA$

and $R_{s min} = \frac{V_{1}}{I_{s max.}} = \frac{3}{7} k Ω$

Similarly, $I_{s min.} = 5 mA$

and $R_{s max.} = \frac{V_{1}}{I_{s min.}} = \frac{3}{5} k Ω$

$∴$ $\frac{3}{7} k Ω < R_{s} < \frac{3}{5} k Ω$

Solution

Q.
In the given circuit if the power rating of Zener diode is 10 mW, the value of series resistance $R_{s}$ to regulate the input unregulated supply is [2024]

1 0.5 $kΩ$

2 10 $Ω$

3 1 $kΩ$

4 10 $kΩ$

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Solution

Q. In the given circuit if the power rating of Zener diode is 10 mW, the value of series resistance Rs to regulate the input unregulated supply is [2024]

1 0.5 kΩ

2 10 Ω

3 1 kΩ