You are again working with Ohm's Law, which is:

\[
V = I \times R
\]

Given:
- \( I = 30 \, \text{mA} \) (which is \( 0.030 \, \text{A} \))
- \( R = 1.8 \, \text{k}\Omega \) (which is \( 1800 \, \Omega \))

Now, substitute the values into the equation:

\[
V = 0.030 \, \text{A} \times 1800 \, \Omega = 54 \, \text{V}
\]

So, the voltage \( V \) is \( 54 \) volts.

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the voltage change if the current were increased to 50 mA?
2. What would the current be if the voltage were 100 V instead?
3. How can you convert resistance from \( \Omega \) to \( k\Omega \)?
4. If the resistance is halved, what will be the new voltage?
5. How does the power dissipated in the resistor depend on the current?

**Tip:** When working with milliamps and kilohms, always convert to base units (amps and ohms) before using Ohm's Law for consistency.

Learn how to calculate voltage using Ohm's Law with a detailed example. Given a current of 30 mA and a resistance of 1.8 kΩ, find the voltage using V = I * R. This problem explores electrical circuits and practical applications of Ohm's Law. Suitable for advanced high school students and enthusiasts.

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