A backup telemetry transmitter operates from a regulated 60\text{ V} direct-current (DC) p… — Armed Services Vocational Aptitude Battery (ASVAB)

A backup telemetry transmitter operates from a regulated $60\text{ V}$ direct-current (DC) power source connected across two parallel branch circuits. The primary transmitter branch has a resistance of $15\ \Omega$ , and the secondary cooling fan branch has a resistance of $30\ \Omega$ . What is the total current drawn from the power source?

A
$8\text{ A}$
$6\text{ A}$ Answer
C
$1.33\text{ A}$
D
$10\text{ A}$

Answer

The total current drawn from the power source is

6\text{ A}

The correct answer is

6\text{ A}

. In a parallel circuit, each branch experiences the full source voltage of

60\text{ V}

. Applying Ohm's Law to each branch yields

I_1 = \frac{60\text{ V}}{15\ \Omega} = 4\text{ A}

and

I_2 = \frac{60\text{ V}}{30\ \Omega} = 2\text{ A}

. Summing these branch currents gives a total current of

4\text{ A} + 2\text{ A} = 6\text{ A}

. Alternatively, finding the equivalent resistance (

R_{\text{eq}} = \frac{15 \times 30}{15 + 30} = 10\ \Omega

) and applying Ohm's Law (

I_{\text{total}} = \frac{60\text{ V}}{10\ \Omega}

) confirms

6\text{ A}

Step-by-Step Solution

Determine the voltage across each parallel branch

V_1 = V_2 = 60\text{ V}

In a parallel circuit, the full supply voltage drops equally across all parallel branches.

Calculate the individual branch currents using Ohm's Law (

I = \frac{V}{R}

)

I_1 = \frac{60\text{ V}}{15\ \Omega} = 4\text{ A}

I_2 = \frac{60\text{ V}}{30\ \Omega} = 2\text{ A}

Each branch current is determined independently by the branch resistance and the common voltage.

Sum the branch currents to find the total circuit current (

I_{\text{total}} = I_1 + I_2

)

I_{\text{total}} = 4\text{ A} + 2\text{ A} = 6\text{ A}

According to Kirchhoff's Current Law, total current entering a parallel junction equals the sum of currents leaving through the branches.

Key Concept

In a DC parallel circuit, voltage remains constant across all branches, and the total current supplied is the sum of the individual branch currents (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

Alternative Method

Find equivalent resistance first:

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{15 \times 30}{15 + 30} = \frac{450}{45} = 10\ \Omega

. Then calculate total current:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{60\text{ V}}{10\ \Omega} = 6\text{ A}

Estimated Time:1m 0s

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