Question

(g) 3 to pass 25 ma of current through a 300 - kω lamp, what voltage is applied across the lamp?

Explanation:

Step1: Recall Ohm's Law

Ohm's Law states that \( V = I \times R \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.

Step2: Convert units

First, convert the current from milliamperes (mA) to amperes (A) and resistance from kilohms (kΩ) to ohms (Ω).

• Current: \( 25 \, \text{mA} = 25 \times 10^{-3} \, \text{A} = 0.025 \, \text{A} \)
• Resistance: \( 300 \, \text{kΩ} = 300 \times 10^{3} \, \Omega = 300000 \, \Omega \)

Step3: Apply Ohm's Law

Substitute \( I = 0.025 \, \text{A} \) and \( R = 300000 \, \Omega \) into the formula \( V = I \times R \).
\( V = 0.025 \, \text{A} \times 300000 \, \Omega \)
\( V = 7500 \, \text{V} \) or \( 7.5 \, \text{kV} \)

Answer:

The voltage applied across the lamp is \( 7500 \, \text{V} \) (or \( 7.5 \, \text{kV} \)).