An electric bulb of 500 W at 100 V is used in a circuit having a 200 V supply. Calculate the resistance $$R$$ to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

We start by recalling the basic power relation for any electrical element:

$$P = V\,I$$

The electric bulb is rated at 500 W when the potential difference across it is 100 V. Substituting these rated values in the formula gives the current that must flow through the bulb for it to operate at its specified power.

$$I = \frac{P}{V} = \frac{500\ \text{W}}{100\ \text{V}} = 5\ \text{A}$$

So, the bulb needs a current of 5 A.

Next, we calculate the resistance of the bulb itself using Ohm’s law in its power form:

First state the relation connecting power, voltage and resistance:

$$P = \frac{V^{2}}{R}$$

Rearranging gives

$$R = \frac{V^{2}}{P}$$

Substituting the rated values of the bulb, we obtain

$$R_{\text{bulb}} = \frac{(100\ \text{V})^{2}}{500\ \text{W}} = \frac{10000\ \text{V}^{2}}{500\ \text{W}} = 20\ \Omega$$

Now the circuit is to be powered from a 200 V supply while the bulb must still carry only 5 A. Because the bulb and the extra resistance $$R$$ are to be connected in series, the same current of 5 A will pass through both.

Using Ohm’s law $$V = I R$$ for the whole series combination, the total resistance required for a 200 V supply and 5 A current is

$$R_{\text{total}} = \frac{V_{\text{supply}}}{I} = \frac{200\ \text{V}}{5\ \text{A}} = 40\ \Omega$$

The total resistance is the sum of the bulb’s own resistance and the external series resistance:

$$R_{\text{total}} = R_{\text{bulb}} + R$$

Substituting the known values,

$$40\ \Omega = 20\ \Omega + R$$

Solving for $$R$$ gives

$$R = 40\ \Omega - 20\ \Omega = 20\ \Omega$$

Thus, a series resistance of 20 Ω must be connected with the bulb so that the bulb still dissipates 500 W when the circuit is operated from a 200 V supply.

Hence, the correct answer is Option C.