A direct current circuit is powered by a 48\text{ V} source. The circuit contains a main l… — Armed Services Vocational Aptitude Battery (ASVAB)

A direct current circuit is powered by a $48\text{ V}$ source. The circuit contains a main line with resistor $R_1$ ( $8\text{ }\Omega$ ) which then splits into two parallel branches containing resistor $R_2$ ( $12\text{ }\Omega$ ) and resistor $R_3$ ( $24\text{ }\Omega$ ) respectively. How much electrical power is dissipated specifically by resistor $R_2$ ?

$48\text{ W}$ Answer
B
$72\text{ W}$
C
$108\text{ W}$
D
$192\text{ W}$

Answer

The power dissipated specifically by resistor

R_2

48\text{ W}

To find the power dissipated by

R_2

, we must determine either the specific voltage across it or the specific current through it. First, we find the parallel equivalent resistance of

R_2

and

R_3

12\text{ }\Omega

and

24\text{ }\Omega

in parallel give

8\text{ }\Omega

. Adding

R_1

(

8\text{ }\Omega

) gives a total circuit resistance of

16\text{ }\Omega

. The total circuit current is

48\text{ V} / 16\text{ }\Omega = 3\text{ A}

. This

3\text{ A}

current flows through the parallel equivalent resistance (

8\text{ }\Omega

), creating a voltage drop of

24\text{ V}

across the parallel branches. Using this specific voltage for

R_2

, we calculate its power dissipation:

P = V^2 / R = 24^2 / 12 = 576 / 12 = 48\text{ W}

Step-by-Step Solution

Calculate the equivalent resistance of the parallel network containing

R_2

and

R_3

R_p = \frac{12 \times 24}{12 + 24} = \frac{288}{36} = 8\text{ }\Omega

This condenses the parallel branches into a single equivalent resistance to find the total circuit resistance.

Calculate the total resistance of the circuit.

R_{total} = R_1 + R_p = 8 + 8 = 16\text{ }\Omega

Adding the series resistor to the equivalent parallel resistance gives the total resistance.

Calculate the total current flowing from the source.

I_{total} = \frac{V_{total}}{R_{total}} = \frac{48}{16} = 3\text{ A}

Ohm's Law allows us to find the total current, which flows entirely through

R_1

Calculate the voltage drop across the parallel network.

V_p = I_{total} \times R_p = 3 \times 8 = 24\text{ V}

Since components in parallel share the same voltage, this is the voltage across both

R_2

and

R_3

Calculate the power dissipated by

R_2

P_{R2} = \frac{V_p^2}{R_2} = \frac{24^2}{12} = \frac{576}{12} = 48\text{ W}

Using Watt's Law with the specific voltage across

R_2

and its specific resistance gives its power dissipation.

Key Concept

Calculating power in a series-parallel circuit requires identifying the specific voltage across or current through the individual component, rather than using total circuit values.

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