For the circuit shown in FIGURE P28.58, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

We need to find the current through and the potential difference across each resistor.

When the resistors are connected one by one and without any junction between them then they are called to be connected in series. For a series of resistors, the equivalent resistance for their combination is given by an equation in the form

$R_{eq}=R_1+R_2+....+R_N\left(1\right)$

The two resistors $5\Omega$and $10\Omega$are in series. Using equation $\left(1\right)$to get their equivalent resistance by

localid="1648959548318" $R_{1,eq}=5\Omega +10\Omega =15\Omega$

The current is the same for resistors in series. So the current through both resistors is calculated by using Ohm's law

$I_5=I_{10}=\frac{\epsilon }{R_{1,eq}}=\frac{24V}{15\Omega }=1.6A$

Use Ohm's law to get the potential difference across each resistor of them by

localid="1648959842005" $△V_5=I_5{R}_5=(1.6A)\left(5\Omega \right)=8V$

And

$△V_{10}=I_{10}{R}_{10}=(1.6A)\left(10\Omega \right)=16V$

The two resistors $4\Omega and6\Omega$are in series. so, equation $\left(1\right)$to get their equivalent resistance by

$R_{2,eq}=6\Omega +4\Omega =10\Omega$

The current is the same for resistance in series, so the current through both the resistance is calculated by using Ohm's law

role="math" localid="1648960283489" $I_6=I_4=\frac{\epsilon }{R_{2,eq}}=\frac{24V}{10\Omega }=2.4A$

Now, use Ohm's law to get the potential diffrence across each rsistor of them by,

$△V_6=I_6{R}_6=(2.4A)\left(6\Omega \right)=14.4V$

And

$△V_4=I_4{R}_4=(2.4A)\left(4\Omega \right)=9.6V$

Summerize the results in the next table