An industrial control panel utilizes a direct-current bus to power three parallel branches… — Armed Services Vocational Aptitude Battery (ASVAB)

An industrial control panel utilizes a direct-current bus to power three parallel branches. Branch 1 contains a $15\ \Omega$ resistor, Branch 2 contains a $30\ \Omega$ resistor, and Branch 3 contains a $10\ \Omega$ resistor. If an ammeter connected in Branch 2 measures a current of $3\text{ A}$ , what is the total current delivered by the power supply to the entire circuit?

A
$4.5\text{ A}$
B
$6.0\text{ A}$
C
$9.0\text{ A}$
$18.0\text{ A}$ Answer

Answer

18.0\text{ A}

The supply voltage across the parallel circuit is constant. Using Branch 2,

V = 3\text{ A} \times 30\ \Omega = 90\text{ V}

. The remaining branch currents are

I_1 = 90\text{ V} / 15\ \Omega = 6\text{ A}

and

I_3 = 90\text{ V} / 10\ \Omega = 9\text{ A}

. Summing all three branch currents yields

I_{\text{total}} = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

Step-by-Step Solution

Calculate the common supply voltage across the parallel circuit using Ohm's Law on Branch 2.

V = I_2 \times R_2 = 3\text{ A} \times 30\ \Omega = 90\text{ V}

In a parallel circuit, the voltage drop across every individual branch is equal to the total supply voltage.

Determine the current flowing through Branch 1 and Branch 3 using the common voltage of

90\text{ V}

I_1 = \frac{90\text{ V}}{15\ \Omega} = 6\text{ A}

and

I_3 = \frac{90\text{ V}}{10\ \Omega} = 9\text{ A}

Current in each individual branch is determined by dividing the constant branch voltage by that branch's resistance.

Apply Kirchhoff's Current Law to sum the currents of all parallel branches.

I_{\text{total}} = I_1 + I_2 + I_3 = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

The total current leaving the power supply equals the sum of the individual branch currents in a parallel circuit.

Key Concept

In a parallel circuit, voltage remains constant across all branches (

V_{\text{total}} = V_1 = V_2 = V_3

), and total circuit current is the sum of all individual branch currents (

I_{\text{total}} = I_1 + I_2 + I_3

Alternative Method

Calculate total equivalent resistance first:

\frac{1}{R_{\text{eq}}} = \frac{1}{15\ \Omega} + \frac{1}{30\ \Omega} + \frac{1}{10\ \Omega} = \frac{2 + 1 + 3}{30} = \frac{6}{30} = \frac{1}{5\ \Omega}

, giving

R_{\text{eq}} = 5\ \Omega

. With

V = 90\text{ V}

, total current is

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{90\text{ V}}{5\ \Omega} = 18\text{ A}

Estimated Time:1m 30s

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