91Ó°ÊÓ

• 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

a)

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

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

$$\begin{gathered} \frac{1}{R_{\text{eq}}}=\frac{1}{R_{1.6}}+\frac{1}{R_{2.4}}+\frac{1}{R_{4.8}} \\ =\frac{1}{1.6\mathrm{Î\copyright }}+\frac{1}{2.4\mathrm{Î\copyright }}+\frac{1}{4.8\mathrm{Î\copyright }} \\ =\frac{5}{4\mathrm{Î\copyright }} \\ {R}_{\text{eq}}=0.8\mathrm{Î\copyright } \end{gathered}$$

The equivalent resistance of the combination is 0.8 Ω.

b)

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

$\mathbf{I}\mathbf{=}\frac{\mathbf{v}}{\stackrel{\mathbf{¯}}{\mathbf{R}}}$

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

The three resistors are connected in parallel. This means the current is not the same for the three resistors and we could use Ohm's law to get the current for each resistor, where the voltage is the same for the three resistors. The current can be calculated as:

Current flowing through R₁ can be calculated as:

$\begin{array}{r}{I}_{1.6}=\frac{28\mathrm{V}}{1.6\mathrm{Î\copyright }}\\ =17.50\mathrm{A}\end{array}$

Current flowing through R₂ can be calculated as:

$\begin{array}{r}{I}_{2,4}=\frac{28\mathrm{V}}{2.4\mathrm{Î\copyright }}\\ =11.67\mathrm{A}\end{array}$

Current flowing through R₃ can be calculated as:

$\begin{array}{r}{I}_{4,8}=\frac{28\mathrm{V}}{4.8\mathrm{Î\copyright }}\\ =5.83\mathrm{A}\end{array}$

The current in resistor R₁ is 17.50 A, resistor R₂ is 11.67 A, and resistor R₃ is 5.83 A.

c)

The total current through resistors connected in parallel is the sum of the current through the individual resistors, and we get:

$$\begin{gathered} l_t=l_{1.6}+l_{2.4}+l_{4.8} \\ =35A \end{gathered}$$

The total current through the battery is 35 A.

d)

Since the three resistors are connected in parallel with the battery, therefore, the potential difference across resistors connected in parallel is the same for every resistor and equals the potential difference across the combination.

$$\begin{gathered} \mathrm{Î\mu }=V_{1.6} \\ =V_{2.4} \\ =V_{4.8} \\ =28V \end{gathered}$$

The voltage across each resistor is the same =28 V

e)

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.

Power dissipation in R₁ resistor:

$$\begin{gathered} P_{1.6}=\left(V_{1.6}\right)\left(I_{1.6}\right) \\ =\left(28\mathrm{V}\right)\left(17.50\mathrm{A}\right) \\ =490\mathrm{W} \end{gathered}$$

Power dissipation in R₂ resistor:

$$\begin{gathered} P_{2.4}=\left(V_{2.4}\right)\left(I_{2.4}\right) \\ =\left(28\mathrm{V}\right)(11.67\mathrm{A}) \\ =326\mathrm{W} \end{gathered}$$

Power dissipation in R₃ resistor:

$\begin{array}{r}{P}_{4.8}=\left(V_{4.8}\right)\left(I_{4.8}\right)\\ =\left(28\mathrm{V}\right)\left(5.83\mathrm{A}\right)\\ =163\mathrm{W}\end{array}$

The power dissipated in resistor R₁ is 490 W, resistor R₂is 326 W, and resistor R₃ is 163 W.

f)

The power dissipated is given by:

$\mathit{P}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}$

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

In a circuit where the resistors are connected in parallel, the voltage across each resistor is the same. Since the voltage is fixed, smaller resistance results in higher power dissipation.