Now the three resistors of Exercise 26.8 are connected in series to the same battery. Answer the same questions for this situation.

• Resistor 1, R₁= 1.60 Ω
• Resistor 2, R₂= 2.40 Ω
• Resistor 3, R₃= 4.80 Ω
• Emf of the battery = 28.0 V

The equivalent resistance of resistors connected in series is given by:

${\mathbf{R}}_{\mathbf{eq}}\mathbf{=}{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}\mathbf{+}{\mathbf{R}}_{\mathbf{3}}$

$$\begin{gathered} R_{\text{eq}}=R_{1.6}+R_{2.4}+R_{4.8} \\ =1.6\mathrm{\Omega }+2.4\mathrm{\Omega }+4.8\mathrm{\Omega } \\ =8.8\mathrm{\Omega } \end{gathered}$$

The equivalent resistance of the combination is 8.8 Ω.

According to Ohm’s Law, the current I flowing through the circuit which has a resistor that has resistance R( in this case equivalent resistance) can be written as:

$\mathbf{I}\mathbf{=}\frac{\mathbf{v}}{\overline{\mathbf{R}}}$

Here V is the potential/voltage drop across the resistor.

The three resistors are connected in series. This means the current is the same for both resistance which equals to the current of their combination

$$\begin{gathered} I=\frac{V}{R_{\text{eq}}} \\ =\frac{28.0\mathrm{V}}{8.8\mathrm{\Omega }} \\ =3.18\mathrm{A} \\ ={\mathrm{I}}_{1.6} \\ ={\mathrm{I}}_{2.4} \\ ={\mathrm{I}}_{4.8} \\ =3.18\mathrm{A} \end{gathered}$$

The current across each resistor is =3.18 A.

The total current through the battery could be calculated by knowing the equivalent resistance. Where the current is flow due to the voltage drop occurring due to the equivalent resistance and from Ohm's law we could calculate this current by:

$$\begin{gathered} \begin{array}{ll}{l}_t=\frac{v}{R_{eq}}& \end{array} \\ =3.18A \end{gathered}$$

The total current through the battery is 3.18 A.

The total potential difference across resistors connected in series is the sum of the individual potential differences where each resistor has its own voltage, but the same current flows in the three resistors. So we could use Ohm's law to get V for each resistor as:

$$\begin{gathered} V_{1.8}=I{R}_{1.6} \\ =\left(3.18\mathrm{A}\right)\left(1.6\mathrm{\Omega }\right) \\ =5.09\mathrm{V} \\ {\mathrm{V}}_{2.4}={\mathrm{IR}}_{2.4} \\ =\left(3.18\mathrm{A}\right)\left(2.4\mathrm{\Omega }\right) \\ =7.63\mathrm{V} \\ {\mathrm{V}}_{4.8}={\mathrm{IR}}_{4.8} \\ =\left(3.18\mathrm{A}\right)\left(4.8\mathrm{\Omega }\right) \\ =15.26\mathrm{V} \end{gathered}$$

The voltage across resistor R₁ is 5.09 V, resistor R₂ is 7.63 V, and resistor R₃ is 15.26 V.

The dissipated power in the resistor is related to the voltage V between the terminals of the resistor as given:

P=VI

Here I is the flowing current.

$$\begin{gathered} P_{1.8}=I{V}_{1.8} \\ =\left(3.18\mathrm{A}\right)\left(5.09\mathrm{V}\right) \\ =16.18\mathrm{W} \\ {\mathrm{P}}_{2.4}={\mathrm{IV}}_{2.4} \\ =\left(3.18\mathrm{A}\right)\left(7.63\mathrm{V}\right) \\ =24.27\mathrm{W} \\ {\mathrm{P}}_{4.8}={\mathrm{IV}}_{4.8} \\ =\left(3.18\mathrm{A}\right)\left(15.26\mathrm{V}\right) \\ =48.54\mathrm{W} \end{gathered}$$

The power dissipated in resistor R₁ is 16.18 W, resistor R₂is 24.27 W, and resistor R₃ is 48.54 W.

The power dissipated is given by:

$\mathbf{P}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{R}$

As shown by the results in Step 5, the most dissipated power is due to the greatest resistance.

In a circuit where the resistors are connected in series, the current across each resistor is the same. Since the current is fixed, larger resistance results in higher power dissipation.