A 6 V battery is connected across a resistance R. The current through R is 0.4 mA. What will be the value of R?

Given:
- Voltage, V = 6 V
- Current, I = 0.4 mA

According to Ohm's law, [tex]V = IR[/tex].
The equation for resistance can be written as:
[tex]R = \frac{V}{I} = \frac{6}{0.4 \times 10^{-3}} = 15 \times 10^{3} \, \Omega = 15 \, k\Omega[/tex].

To solve this problem, we use Ohm's Law, which is a fundamental principle in physics relating voltage (V), current (I), and resistance (R) in an electrical circuit. Ohm's Law is given by the formula:

[tex]V = I \times R[/tex]

Where:

[tex]V[/tex] is the voltage across the resistor (measured in volts, V)
[tex]I[/tex] is the current flowing through the resistor (measured in amperes, A)
[tex]R[/tex] is the resistance of the resistor (measured in ohms, Ω)

In this problem, we are given the following values:

Voltage, [tex]V = 6 \text{ V}[/tex]
Current, [tex]I = 0.4 \text{ mA}[/tex]

Before using Ohm's Law to find the resistance, convert the current from milliamperes (mA) to amperes (A). Since there are 1000 mA in an A, we convert [tex]0.4 \text{ mA}[/tex] to amperes:

[tex]I = 0.4 \times 10^{-3} \text{ A}[/tex]

Now, apply Ohm's Law to solve for the resistance [tex]R[/tex]:

[tex]R = \frac{V}{I} = \frac{6 \text{ V}}{0.4 \times 10^{-3} \text{ A}}[/tex]

[tex]R = \frac{6}{0.0004}[/tex]

[tex]R = 15000 \text{ Ω}[/tex]

Convert ohms to kilohms (kΩ) for a more convenient unit:

Since 1 kΩ = 1000 Ω:

[tex]R = 15 \text{ kΩ}[/tex]

Thus, the resistance [tex]R[/tex] of the circuit is [tex]15 \text{ kΩ}[/tex]. This straightforward calculation demonstrates the relationship between voltage, current, and resistance in an electric circuit using Ohm's Law.