A 30V, 40 w lamp is to be operated on a 120 V DC lines. For proper glows, how much resistance in `Omega` should be connected in series with the lamp?

Options

A. `67.5Omega`
B. `4.75Omega`
C. `75.6Omega`
D. none

Answer

To solve the problem of determining the resistance needed to operate a 30V, 40W lamp on a 120V DC line, we can follow these steps:

### Step 1: Calculate the Current Required by the Lamp
The power (P) of the lamp is given as 40W, and the voltage (V) across the lamp is 30V. We can use the formula for power:

\[ P = V \times I \]

Where:
- \( P \) = Power (in watts)
- \( V \) = Voltage (in volts)
- \( I \) = Current (in amperes)

Rearranging the formula to find the current \( I \):

\[ I = \frac{P}{V} \]

Substituting the values:

\[ I = \frac{40W}{30V} = \frac{4}{3} A \]

### Step 2: Determine the Total Voltage and the Voltage Across the Resistor
The total voltage supplied is 120V. The lamp operates at 30V, so the voltage drop across the resistor (R) can be calculated as:

\[ V_R = V_{total} - V_{lamp} \]

Substituting the values:

\[ V_R = 120V - 30V = 90V \]

### Step 3: Calculate the Required Resistance
Using Ohm's Law, we can express the relationship between voltage, current, and resistance:

\[ V = I \times R \]

We can rearrange this to find the resistance \( R \):

\[ R = \frac{V_R}{I} \]

Substituting the values we calculated:

\[ R = \frac{90V}{\frac{4}{3} A} \]

To simplify:

\[ R = 90V \times \frac{3}{4} = \frac{270}{4} = 67.5 \Omega \]

### Conclusion
The resistance that should be connected in series with the lamp is **67.5 Ω**.

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