Question

A student in the laboratory connects a 14 Ω resistor, a 22 Ω resistor, and a 34 Ω resistor in parallel and then connects the arrangement to a 52 V dc source.(a) What is the current? _______:A(b) What is the power? ________:W

Answer 1

Given:

The resistances connected in parallel combination are:

R1 = 14 Ω

R2 = 22 Ω

R3 = 34 Ω

The voltage applied is: V = 52 v dc

To find:

a) The current

b) The power

Explanation:

a)

The equivalent resistance R of all three resistors connected in parallel is given as:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Substituting the values in the above equation, we get:

$\begin{gathered} \frac{1}{R}=\frac{1}{14\text{ \Omega}}+\frac{1}{22\text{ \Omega}}+\frac{1}{34\text{ \Omega}} \\ \\ \frac{1}{R}=\frac{22}{22\times14\text{ \Omega}}+\frac{14}{14\times22\text{ \Omega}}+\frac{1}{34\text{ \Omega}} \\ \\ \frac{1}{R}=\frac{36}{308\text{ \Omega}}+\frac{1}{34\text{ \Omega}} \\ \\ \frac{1}{R}=\frac{36\times34}{34\times308\text{ \Omega}}+\frac{308}{308\times34\text{ \Omega}} \\ \\ \frac{1}{R}=\frac{1224}{10472\text{ \Omega}}+\frac{308}{10472\text{ \Omega}} \\ \\ \frac{1}{R}=\frac{1532}{10472\text{ \Omega}} \\ \\ R=\frac{10472\text{ \Omega}}{1532} \\ \\ R=6.8355\text{ \Omega} \end{gathered}$

Now, this equivalent resistance R of the parallel combination of resistors acts as a single resistance through which the current I passes when 52 v voltage is applied across it. The current can be calculated by using Ohm's law as:

$\begin{gathered} V=IR \\ \\ I=\frac{V}{R} \end{gathered}$

Substituting the values in the above equation, we get:

$I=\frac{52\text{ v}}{6.8355\text{ \Omega}}=7.6073\text{ A}$

b)

The power P is given as:

$P=IV$

Substituting the values in the above equation, we get:

$P=7.6073\text{ A}\times52\text{ v}=395.58\text{ W}$

Final answer:

a) The current is 6.8355 Amperes.

b) The power is 395.58 watts.