Is My Battery Completely Dead?

[Sho Kano]

Homework Statement

When a certain 9.0V battery is connected to a voltmeter, the voltmeter reads 8.00V (assume that the voltmeter has an internal resistance of 10MΩ. We may conclude that
(A) only 1/9 of the battery's fuel has been used, and so the battery has plenty of life left in it.
(B) the battery is nearing the end of its life, but is still usable.
(C) the battery is unable to provide more than a few μA of current to any circuit (even a short circuit), and so is therefore (for all practical purposes) completely dead.

Homework Equations

V = IR

The Attempt at a Solution

No idea honestly. It seems that the battery has plenty of life in it, but that's the wrong answer.

 

[ehild]

Homework Statement

When a certain 9.0V battery is connected to a voltmeter, the voltmeter reads 8.00V (assume that the voltmeter has an internal resistance of 10MΩ. We may conclude that
(A) only 1/9 of the battery's fuel has been used, and so the battery has plenty of life left in it.
(B) the battery is nearing the end of its life, but is still usable.
(C) the battery is unable to provide more than a few μA of current to any circuit (even a short circuit), and so is therefore (for all practical purposes) completely dead.

Homework Equations

V = IR

The Attempt at a Solution

No idea honestly. It seems that the battery has plenty of life in it, but that's the wrong answer.

Usually, a voltmeter has high resistance,more than 1000 ohms, or even much higher. If it reads 8 V across the terminal of a 9 V battery, what can be the internal resistance of the battery in terms of the voltmeter resistance? If you load the battery with a light bulb of about 1 ohm resistance, what will be the voltage across it?

 

[haruspex]

Homework Statement

When a certain 9.0V battery is connected to a voltmeter, the voltmeter reads 8.00V (assume that the voltmeter has an internal resistance of 10MΩ. We may conclude that
(A) only 1/9 of the battery's fuel has been used, and so the battery has plenty of life left in it.
(B) the battery is nearing the end of its life, but is still usable.
(C) the battery is unable to provide more than a few μA of current to any circuit (even a short circuit), and so is therefore (for all practical purposes) completely dead.

Homework Equations

V = IR

The Attempt at a Solution

No idea honestly. It seems that the battery has plenty of life in it, but that's the wrong answer.

What current flowed when connected to the voltmeter? (You can work that out.).

 

[CWatters]

One way to model a battery is to think of it as a constant ideal voltage source of 9V and an internal series resistor. When the battery is fully charged the internal resistance is small. As it runs down the internal resistance gets larger.

Try drawing a battery like this with another resistor connected to represent the voltmeter. How large must the internal resistor be to give 8v on the terminals?

 

[Sho Kano]

What current flowed when connected to the voltmeter? (You can work that out.).

8/10e6 = 8e-7. But what does that say? Of course there will be like no current through the voltmeter.

 

[SammyS]

8/10e6 = 8e-7. But what does that say? Of course there will be like no current through the voltmeter.

Of course the current DOES flow through the voltmeter.

Model the battery as being an ideal voltage source of 9.0 V in series with an internal resistor. Therefore, 1.0 V is dropped across that internal resistor. What is the resistance value of that resistor?

 

[Sho Kano]

Of course the current DOES flow through the voltmeter.

Model the battery as being an ideal voltage source of 9.0 V in series with an internal resistor. Therefore, 1.0 V is dropped across that internal resistor. What is the resistance value of that resistor?

V = IR
1.0 = 8e-7*R
R = 1250000 ohms ?

 

[SammyS]

V = IR
1.0 = 8e-7*R
R = 1250000 ohms ?

Correct.

Now consider using that battery to power a 1Watt light bulb which is designed to be used at 9 V .

 

[Sho Kano]

SammyS,
P = IV
1 = 9*I
I = 0.11 A

We need 0.11 A
We have 8e-7 A
Is that right?

 

[SammyS]

P = IV
1 = 9*I
I = 0.11 A

We need 0.11 A
We have 8e-7 A
Is that right?

That's essentially correct.

At operating conditions, the bulb has a resistance of about 80 Ω . ( P = V²/R )

That corresponds to a voltage drop of 8 or 9 micro-volts and a power output on the order of 1 micro-watt.

 

[Sho Kano]

That's essentially correct.

At operating conditions, the bulb has a resistance of about 80 Ω . ( P = V²/R )

That corresponds to a voltage drop of 8 or 9 micro-volts and a power output on the order of 1 micro-watt.

So because the battery cannot even supply a μA of current to even a short circuit, the battery is completely dead. The answer is (C).