Answer
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Hint: To find the value of a series resistor, we can use the Ohm’s law but first, we will need the current and voltage that will be passing through it. Current passing through it will be the total current of the circuit, i.e. the sum of Zener current and load current. Voltage drop across it will be equal to the difference between the unregulated input and Zener voltage.
Formula Used:
According to Ohm’s Law, Resistance, $R = \dfrac{V}{I}$ where, $V$ is the voltage and $I$ is the current across the circuit.
Complete Step by Step Solution:
First, we will have a look at the values given to us.
We have, Zener voltage, ${V_Z} = 6V$
Load Current, ${I_L} = 4mA = 4 \times {10^{ - 3}}A$ (conversion to SI unit)
Unregulated input $ = 10V$
And Zener Current, ${I_Z} = 5{I_L}$
We have converted the values to SI units as well, wherever required.
Now, the current passing through the series resistance will be equal to the total current of the circuit. i.e. $I = {I_Z} + {I_L}$
We know that ${I_Z} = 5{I_L}$ . Using this equation in the above formula, we get $I = 5{I_L} + {I_L}$
Hence, $I = 6{I_L}$
On putting the values, we get $I = 6 \times 4 \times {10^{ - 3}} = 24 \times {10^{ - 3}}A$
Now we have the value of current through the series resistance and need the value of voltage drop through it which will be equal to the difference between the unregulated input and Zener voltage.
That is, ${V_S} = 10 - 6 = 4V$
Now, using Ohm’s Law, we can get the value of series resistance.
Series resistance, ${R_S} = \dfrac{{{V_S}}}{I}$
On putting the values, we get ${R_S} = \dfrac{4}{{24 \times {{10}^{ - 3}}}}$
On further solving, we get ${R_S} = 0.16 \times {10^3}\Omega $
Or, we can say that ${R_S} = 160\Omega $
Hence, value of ${R_S}$ is greater than $100\Omega.$
Therefore, option (C) is the correct answer.
Note: A Zener diode is used in a number of experiments. It is a silicon semiconductor and allows current to flow through it in forward as well as reverse direction. When current flows in forward direction, it is said to be forward biased and when current flows in reverse direction, it is said to be reverse biased.
Formula Used:
According to Ohm’s Law, Resistance, $R = \dfrac{V}{I}$ where, $V$ is the voltage and $I$ is the current across the circuit.
Complete Step by Step Solution:
First, we will have a look at the values given to us.
We have, Zener voltage, ${V_Z} = 6V$
Load Current, ${I_L} = 4mA = 4 \times {10^{ - 3}}A$ (conversion to SI unit)
Unregulated input $ = 10V$
And Zener Current, ${I_Z} = 5{I_L}$
We have converted the values to SI units as well, wherever required.
Now, the current passing through the series resistance will be equal to the total current of the circuit. i.e. $I = {I_Z} + {I_L}$
We know that ${I_Z} = 5{I_L}$ . Using this equation in the above formula, we get $I = 5{I_L} + {I_L}$
Hence, $I = 6{I_L}$
On putting the values, we get $I = 6 \times 4 \times {10^{ - 3}} = 24 \times {10^{ - 3}}A$
Now we have the value of current through the series resistance and need the value of voltage drop through it which will be equal to the difference between the unregulated input and Zener voltage.
That is, ${V_S} = 10 - 6 = 4V$
Now, using Ohm’s Law, we can get the value of series resistance.
Series resistance, ${R_S} = \dfrac{{{V_S}}}{I}$
On putting the values, we get ${R_S} = \dfrac{4}{{24 \times {{10}^{ - 3}}}}$
On further solving, we get ${R_S} = 0.16 \times {10^3}\Omega $
Or, we can say that ${R_S} = 160\Omega $
Hence, value of ${R_S}$ is greater than $100\Omega.$
Therefore, option (C) is the correct answer.
Note: A Zener diode is used in a number of experiments. It is a silicon semiconductor and allows current to flow through it in forward as well as reverse direction. When current flows in forward direction, it is said to be forward biased and when current flows in reverse direction, it is said to be reverse biased.
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