Whoops, only checked the math cursorily and saw a R1 = 0.514 * RL...Schrodingerscat wrote:
Your numbers are off by I think a factor of 2. Quick check: 5V across 2000 ohms is 2.5mA and across 1800 is 2.8mA, for a total current of 5.3mA across 514.3 ohms, for a voltage of 2.7 volts. Thus you have a 9-5-2.7=1.3V discrepancy in the 5V load configuration. Also finding the maximum efficiency may be too messy for a test (even using a graphing calculator to maximize expressions), although you could find the maximum power to the load using Thevenin equivalent, which I do not believe gives maximum efficiency necessarily (without analyzing this specific configuration).
Shock Value/Circuit Lab Question Marathon
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Re: Circuit Lab Question Marathon
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Re: Circuit Lab Question Marathon
Yep, so did I... saw a blatantly easy mistake in my work. Definitely should be double the values I gave you, R1 = 1028 ohm and R2 = 3600 ohm. The math actually checks out this time, unless I'm making another stupid mistake.killer225whale wrote:Whoops, only checked the math cursorily and saw a R1 = 0.514 * RL...Schrodingerscat wrote:
Your numbers are off by I think a factor of 2. Quick check: 5V across 2000 ohms is 2.5mA and across 1800 is 2.8mA, for a total current of 5.3mA across 514.3 ohms, for a voltage of 2.7 volts. Thus you have a 9-5-2.7=1.3V discrepancy in the 5V load configuration. Also finding the maximum efficiency may be too messy for a test (even using a graphing calculator to maximize expressions), although you could find the maximum power to the load using Thevenin equivalent, which I do not believe gives maximum efficiency necessarily (without analyzing this specific configuration).
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Re: Shock Value/Circuit Lab Question Marathon
Restarting.. (with an easy one because I don't know anything yet) what kind of circuit is a current divider?
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Re: Shock Value/Circuit Lab Question Marathon
Current source in parallel with resistance as well as the branch where you want to divide current.UTF-8 U+6211 U+662F wrote:Restarting.. (with an easy one because I don't know anything yet) what kind of circuit is a current divider?
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Re: Shock Value/Circuit Lab Question Marathon
Yep, your turn.PM2017 wrote:Current source in parallel with resistance as well as the branch where you want to divide current.UTF-8 U+6211 U+662F wrote:Restarting.. (with an easy one because I don't know anything yet) what kind of circuit is a current divider?
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Re: Shock Value/Circuit Lab Question Marathon
Restarting this again
A battery with EMF 8 V is connected to a resistor with resistance 5 kOhms. A voltage of 7.9 V is measured across its terminals. Find the internal resistance of the battery.
A battery with EMF 8 V is connected to a resistor with resistance 5 kOhms. A voltage of 7.9 V is measured across its terminals. Find the internal resistance of the battery.
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Re: Shock Value/Circuit Lab Question Marathon
A internal resistance of 0.1V?
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Re: Shock Value/Circuit Lab Question Marathon
Things2do wrote:Although I think resistance is in Ohms...A internal resistance of 0.1V?
A battery with internal resistance can be modeled as an ideal voltage source connected in series to a resistor. 0.1 V would be the voltage drop across that resistor. Use Ohm's law to find the current in the overall circuit and use it again to find the internal resistance.
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Re: Shock Value/Circuit Lab Question Marathon
Restarting this since it's been a while:
63.3 ohms (although with that kind of internal resistance, you should probably get a new battery)
Let's do some physics. How many electrons flow through a point on a wire carrying 30 mA in 30 seconds?
63.3 ohms (although with that kind of internal resistance, you should probably get a new battery)
Let's do some physics. How many electrons flow through a point on a wire carrying 30 mA in 30 seconds?
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Re: Shock Value/Circuit Lab Question Marathon
mdv2o5 wrote:Restarting this since it's been a while:
63.3 ohms (although with that kind of internal resistance, you should probably get a new battery)
Let's do some physics. How many electrons flow through a point on a wire carrying 30 mA in 30 seconds?
30 mA * 1 A/1000 mA * 30 seconds * 6.24E18 electrons/C = 5.6E18 electrons
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