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(C) The power dissipation $P$ in the protective resistance $R$ is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the resisto

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$0-12 \,V$

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(C) The power dissipation $P$ in the protective resistance $R$ is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the resistor.To determine the minimum voltage range of the $DC$ source required to test the diode up to its maximum power rating,we use the power formula $P = \frac{V^2}{R}$.Given $P = 1 \,W$ and $R = 100 \,\Omega$,we have $V^2 = P 	imes R$.$V^2 = 1 \,W 	imes 100 \,\Omega = 100 \,V^2$.Taking the square root,we get $V = 10 \,V$.Therefore,the minimum voltage range of the $DC$ source must be at least $0-10 \,V$. Among the given options,$0-12 \,V$ is the appropriate range that covers the required $10 \,V$ limit.

An experiment is performed to determine the $I-V$ characteristics of a Zener diode,which has a protective resistance of $R = 100 \,\Omega$ and a maximum power dissipation rating of $P = 1 \,W$. The minimum voltage range of the $DC$ source in the circuit is

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