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Can anyone solve this internal resistance question?
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Fortox
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A battery is connected to a 10 Ω resistor and a switch in series. A voltmeter is connected across the battery. When the switch is open (off) the voltmeter reads 1.45 V. When the switch is closed the reading is 1.26 V.
What is the internal resistance of the battery?
What is the internal resistance of the battery?
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jalili
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(Original post by Fortox)
A battery is connected to a 10 Ω resistor and a switch in series. A voltmeter is connected across the battery. When the switch is open (off) the voltmeter reads 1.45 V. When the switch is closed the reading is 1.26 V.
What is the internal resistance of the battery?
A battery is connected to a 10 Ω resistor and a switch in series. A voltmeter is connected across the battery. When the switch is open (off) the voltmeter reads 1.45 V. When the switch is closed the reading is 1.26 V.
What is the internal resistance of the battery?
Current with switch closed = 1.45/(R+10)
That current flowing in R drops (1.45 - 1.26) volts = 0.19v
R*1.45/(R+10) = 0.19
solve for R, R=1.508
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Fortox
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(Original post by jalili)
Internal resistance = R
Current with switch closed = 1.45/(R+10)
That current flowing in R drops (1.45 - 1.26) volts = 0.19v
R*1.45/(R+10) = 0.19
solve for R, R=1.508
Internal resistance = R
Current with switch closed = 1.45/(R+10)
That current flowing in R drops (1.45 - 1.26) volts = 0.19v
R*1.45/(R+10) = 0.19
solve for R, R=1.508
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uberteknik
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#4
(Original post by Fortox)
By the way, when the switch is open how does current flow at all? If I understand correctly, this circuit would be the same as a voltmeter connected to a battery in series (not in parallel since there is nowhere else for the current to go) and voltmeters have infinite resistance.
By the way, when the switch is open how does current flow at all? If I understand correctly, this circuit would be the same as a voltmeter connected to a battery in series (not in parallel since there is nowhere else for the current to go) and voltmeters have infinite resistance.
In the real world, all voltmeters have a finite resistance - albeit a very high finite resistance (>10x106 Ohms) is typical.
Current still flows through the voltmeter but will be extremely small as a result, <100nA and as low as <10pA for the best (read very expensive) models.
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AndrewSr
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Current with switched closed = 1.26/(R 10)...R*1.26/(R 10) = 0.19Solve for R, R = 1.78
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yosotros
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#6
When the switch is open, the reading is 1.45V. Because the voltmeter theoretically has infinite resistance, there is negligible current flowing across the cell. therefore negligible resistance lost across the internal resistance of the cell. So the emf of the cell is 1.45V. when the switch is closed it reads 1.26V. So the voltage lost across the cell is 1.45-1.26 = 0.19V. 1.26V is lost across the 10 ohm resistor. By applying V=IR to the 10 ohm resistor we can find the current in the circuit. V=IR... I=V/R. = 1.26/10 = 0.126A.Now, across the cell, we have the voltage lost (0.19V), and the current from our previous calculation (0.126A). So now we can apply V=IR to find the resistance across the cell. V=IR... R=V/I. = 0.19/0.126 = 1.508 ohms.
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