A military field radio power pack contains four 3.6\text{ V} direct-current battery cells … — Armed Services Vocational Aptitude Battery (ASVAB)

A military field radio power pack contains four $3.6\text{ V}$ direct-current battery cells connected in series. During maintenance, three cells are placed in series-aiding polarity, but one cell is accidentally reversed into a series-opposing polarity. The power pack is also connected in series with an auxiliary charging module that supplies an aiding electromotive force (EMF) of $850\text{ mV}$ and a protective component that produces an opposing voltage drop of $650\text{ mV}$ . What is the net electromotive force, in volts ( $\text{V}$ ), delivered across the radio terminals?

Answer: 7.4 V

Answer

The net electromotive force delivered across the radio terminals is

7.4\text{ V}

In a series circuit, total electromotive force is the algebraic sum of all individual aiding and opposing potentials. The three aiding

3.6\text{ V}

cells contribute

+10.8\text{ V}

, while the reversed cell subtracts

-3.6\text{ V}

, leaving

+7.2\text{ V}

. Converting the auxiliary units (

850\text{ mV} = +0.85\text{ V}

and

650\text{ mV} = -0.65\text{ V}

) yields a net auxiliary contribution of

+0.20\text{ V}

. Adding these gives a total net terminal EMF of

7.2\text{ V} + 0.20\text{ V} = 7.4\text{ V}

Step-by-Step Solution

Determine the net voltage contribution from the battery cells

Net battery EMF =

7.2\text{ V}

When cells are connected in series, aiding voltages add while opposing voltages subtract. With three cells aiding and one cell opposing:

V_{\text{cells}} = (3.6\text{ V} + 3.6\text{ V} + 3.6\text{ V}) - 3.6\text{ V} = 7.2\text{ V}

Convert auxiliary voltage values from millivolts to volts

Auxiliary aiding EMF =

0.85\text{ V}

, opposing drop =

0.65\text{ V}

Since

1\text{ V} = 1000\text{ mV}

, divide millivolts by

1000

850\text{ mV} = 0.85\text{ V}

and

650\text{ mV} = 0.65\text{ V}

Calculate the net contribution from the auxiliary components

Net auxiliary voltage =

+0.20\text{ V}

Subtract the opposing voltage drop from the aiding auxiliary EMF:

0.85\text{ V} - 0.65\text{ V} = +0.20\text{ V}

Combine the net battery EMF and net auxiliary voltage

Total EMF =

7.4\text{ V}

Add the net auxiliary contribution to the net battery bank voltage:

7.2\text{ V} + 0.20\text{ V} = 7.4\text{ V}

Key Concept

Series Electromotive Force (EMF) Addition, Opposing Polarity, and Metric Prefix Conversion

Estimated Time:1m 30s

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