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**Hint:**By applying Kirchhoff’s Voltage law and using the Power formula, substitute the given values in these relations and to find the minimum voltage range DC source. Electric Power: The electric power in the circuit is the rate at which energy is absorbed or produced within a circuit. It is the product of voltage times current.

**Formula Used:**

We will be using the formula of Kirchhoff’s voltage law and Power $P = VI$.

**Complete step by step solution:**

The circuit diode for a Zener diode with a protective resistance is $R = 100\Omega $.

According to Kirchhoff’s voltage law, the voltage around the loop equals the sum of every voltage drop in the same loop for any closed network and also equals zero.

$V = iR + {V_D}$

Given: The value of the protective resistance is given as $R = 100\Omega $ the power dissipated is $1W$${V_D} = V - iR = V - 100i$

Maximum dissipated across the Zener diode is given by

$P = {V_D}I = (V - 100i)i = 1$

The equation above is equated to zero because Power is given by $1$

Therefore, $100{i^2} - Vi + 1 \geqslant 0$

$i = \dfrac{{V \pm \sqrt {{V^2} - 400} }}{{200}}$

Current should be real, hence the determinant is greater than zero.

${V^2} > 400$

$V > 20V$

The minimum voltage range of the DC source in the circuit is $0 - 24V$

**Hence, Option \[(A)\], $0 - 24V$ is the correct answer.**

**Note:**Kirchhoff’s Voltage Law also called Kirchhoff’s loop rule. In $1845$, a German physicist Gustav Kirchhoff developed a pair of laws with the conservation of current and energy within electrical circuits. In Kirchhoff’s Voltage Law, the voltage drops in all directions either negative or positive and returns to the same point. It is very important to maintain the direction either clockwise or counterclockwise, otherwise, the final voltage will not be equal to zero.

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