A Zener diode with a breakdown voltage of 4 V is connected in series with a resistance $$R$$ to a battery of emf 10 V. The maximum power dissipation rating for the Zener diode is 1 W. The value of $$R$$ to ensure maximum power dissipation across the diode is

We have a Zener diode whose breakdown (Zener) voltage is given as $$V_Z = 4\ \text{V}$$. It is rated for a maximum power dissipation of $$P_{\text{max}} = 1\ \text{W}$$. The diode is connected in series with a resistor $$R$$ to a battery of emf $$E = 10\ \text{V}$$. Our task is to choose $$R$$ so that the diode just reaches, but does not exceed, its maximum power rating.

First, recall the basic power formula that links power, voltage and current:

$$P = V \times I$$

For the Zener diode, the maximum current that can flow when it is at its breakdown voltage without exceeding the rated power is obtained by rearranging the power formula:

$$I_{\text{max}} = \frac{P_{\text{max}}}{V_Z}$$

Substituting the numerical values, we get

$$I_{\text{max}} = \frac{1\ \text{W}}{4\ \text{V}} = 0.25\ \text{A}$$

Now, when the diode is in breakdown, the battery voltage $$E$$ is divided between the resistor $$R$$ and the diode. The voltage across the resistor is therefore

$$V_R = E - V_Z = 10\ \text{V} - 4\ \text{V} = 6\ \text{V}$$

By Ohm’s law, which states $$V = I R$$, the value of the resistor required to sustain the current $$I_{\text{max}}$$ while maintaining this voltage drop is

$$R = \frac{V_R}{I_{\text{max}}}$$

Substituting $$V_R = 6\ \text{V}$$ and $$I_{\text{max}} = 0.25\ \text{A}$$, we obtain

$$R = \frac{6\ \text{V}}{0.25\ \text{A}} = 24\ \Omega$$

Thus the resistor must be $$24\ \Omega$$ to ensure that the Zener diode dissipates exactly its maximum rated power of 1 W when in breakdown.

Hence, the correct answer is Option B.