A `20 V` 5 watt lamp is used in ac main `220 V` and frequency 50 c.p.s.
What pure resistance should be included in place of the above passive elements so that the lamp can run on its rated voltage?

Options

A. `120 Omega`
B. `240 Omega`
C. `800 Omega`
D. `720 Omega`

Answer

To solve the problem of determining the pure resistance that should be included in the circuit with a 20V, 5W lamp when connected to a 220V AC supply, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Current Required by the Lamp:**
The power rating (P) of the lamp is 5W and the rated voltage (V) is 20V. We can calculate the current (I) using the formula:
\[ I = \frac{P}{V} \]
Substituting the values:
\[ I = \frac{5W}{20V} = 0.25A \]

2. **Calculate the Resistance of the Lamp (RL):**
Using Ohm's law, we can find the resistance of the lamp (RL) using the formula:
\[ R_L = \frac{V}{I} \]
Substituting the rated voltage and current:
\[ R_L = \frac{20V}{0.25A} = 80 \, \Omega \]

3. **Determine the Total Resistance Required (R_net):**
The total voltage supplied is 220V, and we want the current to remain at 0.25A. We can calculate the total resistance required (R_net) using Ohm's law:
\[ R_{net} = \frac{V_{applied}}{I} \]
Substituting the values:
\[ R_{net} = \frac{220V}{0.25A} = 880 \, \Omega \]

4. **Calculate the Required Resistance (R):**
The total resistance in the circuit is the sum of the resistance of the lamp and the additional resistance (R) we need to add:
\[ R_{net} = R + R_L \]
Rearranging this gives us:
\[ R = R_{net} - R_L \]
Substituting the values we calculated:
\[ R = 880 \, \Omega - 80 \, \Omega = 800 \, \Omega \]

### Final Answer:
The pure resistance that should be included in the circuit is **800 Ω**.

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