A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 resistor connected in series, the zener current is found to be 4 times load current. The load current and load resistance are: [2025]

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Q.
A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V.

For a 400 $Ω$ resistor connected in series, the zener current is found to be 4 times load current. The load current $(I_{L})$ and load resistance $(R_{L})$ are: [2025]

1 $I_{L} = 20 mA; R_{L} = 250 Ω$

2 $I_{L} = 10 A; R_{L} = 0.5 Ω$

3 $I_{L} = 0.02 mA; R_{L} = 250 Ω$

4 $I_{L} = 10 mA; R_{L} = 500 Ω$

Ans.
(4)

$5 I_{L} = \frac{(25 - 5)}{400} = \frac{1}{20} = 50 mA$

$I_{L} = 10 mA$

$5 = I_{L} R_{L}$

$R_{L} = \frac{5}{I_{L}} = \frac{5}{10 \times 10^{- 3}} = 500 Ω$

Solution

Q.
A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V.

For a 400 $Ω$ resistor connected in series, the zener current is found to be 4 times load current. The load current $(I_{L})$ and load resistance $(R_{L})$ are: [2025]

1 $I_{L} = 20 mA; R_{L} = 250 Ω$

2 $I_{L} = 10 A; R_{L} = 0.5 Ω$

3 $I_{L} = 0.02 mA; R_{L} = 250 Ω$

4 $I_{L} = 10 mA; R_{L} = 500 Ω$

Ans.
(4)

$5 I_{L} = \frac{(25 - 5)}{400} = \frac{1}{20} = 50 mA$

$I_{L} = 10 mA$

$5 = I_{L} R_{L}$

$R_{L} = \frac{5}{I_{L}} = \frac{5}{10 \times 10^{- 3}} = 500 Ω$

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Solution

Q. A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 Ω resistor connected in series, the zener current is found to be 4 times load current. The load current (IL) and load resistance (RL) are: [2025]

1 IL=20 mA; RL=250 Ω

2 IL=10 A; RL=0.5 Ω

3 IL=0.02 mA; RL=250 Ω

4 IL=10 mA; RL=500 Ω

Ans. (4) 5IL=(25-5)400=120=50 mA IL=10 mA 5=ILRL RL=5IL=510×10-3=500 Ω

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Q.
A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V.

For a 400 $Ω$ resistor connected in series, the zener current is found to be 4 times load current. The load current $(I_{L})$ and load resistance $(R_{L})$ are: [2025]

1 $I_{L} = 20 mA; R_{L} = 250 Ω$

2 $I_{L} = 10 A; R_{L} = 0.5 Ω$

3 $I_{L} = 0.02 mA; R_{L} = 250 Ω$

4 $I_{L} = 10 mA; R_{L} = 500 Ω$

Ans.
(4)

$5 I_{L} = \frac{(25 - 5)}{400} = \frac{1}{20} = 50 mA$

$I_{L} = 10 mA$

$5 = I_{L} R_{L}$

$R_{L} = \frac{5}{I_{L}} = \frac{5}{10 \times 10^{- 3}} = 500 Ω$