Question

Four resistors of $3\Omega$, $5\Omega$, $7\Omega$ and $10\Omega$ are connected in series to a $10V$ battery. Calculate the current in $5\Omega$ resistor.

Answer 1

Step 1 - Given Data

The first resistor's resistance is $R_1=3\Omega$

The second resistor's resistance is $R_2=5\Omega$

The third resistor's resistance is $R_3=7\Omega$

And the final resistor's resistance is $R_4=10\Omega$

The voltage of the battery is $V=10V$

Step 2 - Finding equivalent resistance of series combination.

Resistances, $3\Omega$, $5\Omega$, $7\Omega$ , and $10\Omega$, are connected in series. Now let's assume the equivalent resistance for those resistances is $R_{eq}$.

So, $R_{eq}=R_1+R_2+R_3+R_4$

Substitute the values in the equation of $R_{eq}$ to calculate the equivalent resistance.

$$\begin{gathered} R_{eq}=3\Omega +5\Omega +7\Omega +10\Omega \\ {R}_{eq}=25\Omega \end{gathered}$$

Step 3 - Finding the current through the $5\Omega$ resistor.

According to Ohm's law if $i$ is the current through the circuit, $R_{eq}$ is the equivalent resistance of the circuit and $V$ is the voltage applied across the circuit then the relation between them is given by, $i=\frac{V}{R_{eq}}$.

Now substitute the values in the formula of $i$ to calculate the current through the circuit.

$$\begin{gathered} i=\frac{10V}{25\Omega } \\ i=0.4A \end{gathered}$$

As one of the characteristics of series combination of resistances is current through all the resistances is same. This means the supplied current will be the current through the $5\Omega$ resistor.

Final answer - Thus the current through the $5\Omega$ resistance is $0.4A$.