Question

In connection with the circuit drawn below, the value of current flowing through the $2\mathrm{k\Omega }$ resistor is _______ $\times {10}^{-4}\mathrm{A}$.

Answer 1

Explanation:

Step 1. Given data

Voltage in the circuit is = $5\mathrm{V}$ and

Resistance in the circuit is = $2\mathrm{k\Omega }$

We have to find the current through $2\mathrm{k\Omega }$ resistor.

Step 2. Formula to be used

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage.

$$\begin{gathered} \mathrm{V}=\mathrm{IR} \\ \mathrm{Can}\mathrm{be}\mathrm{rewritten}\mathrm{as}, \\ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \end{gathered}$$

Here, $\mathrm{I}$ is current in amperes, $\mathrm{R}$ is resistance in Ohms, and $\mathrm{V}$ is voltage in volts

Step 3. Determine the current.

To find the current, we have to convert the resistor in ohm.

So, $1\mathrm{k\Omega }=1000\mathrm{\Omega }$

Therefore,

$2\mathrm{k\Omega }=2000\mathrm{\Omega }$

$=2\times {10}^3\mathrm{\Omega }$

Now,

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

Substitute the given values in above formula, we get,

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

$=\frac{5\mathrm{V}}{2\times {10}^3\mathrm{\Omega }}$

$=2.5\times {10}^{-3}\mathrm{A}$

$=25\times {10}^{-4}\mathrm{A}$

Hence, in connection with the circuit drawn below, the value of current flowing through the $2\mathrm{k\Omega }$ resistor is $25\times {10}^{-4}\mathrm{A}$.