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A typical car battery can be modeled as an ideal source E connected in series with an internal resistance Rbatt. Assume you have a good battery for which Egood = 12.0 V and Rbatt, good = 0.0200 Ω and a dead one for which Edead = 10.8 V and Rbatt, dead = 0.300 Ω .

Calculate the current through the batteries connected in series.
Calculate the current through the batteries connected in parallel.

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Answer with Explanation:

We are given that

Egood=12 V

Rbatt,good=0.02ohm

Edead=10.8 V

Rbatt,dead=0.3 ohm

We have to find the current through the batteries connected in series and parallel.

When batteries are connected in series

V=Egood+Edead=12+10.8=22.8 V

R=Rbatt,good+Rbatt,dead=0.02+0.3=0.32 ohm

Current,I=[tex]\frac{V}{R}[/tex]

Using the formula

[tex]I=\frac{22.8}{0.32}=71.25 A[/tex]

When the  batteries are connected in parallel

V=Egood-Edead=12-10.8=1.2 V

R=0.02+0.3=0.32 ohm

[tex]I=\frac{1.2}{0.32}=3.75 A[/tex]

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