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A cell of emf 3.0 volts and internal resistance of 0.01 ohm is connected through an ammeter of 0.05-ohm internal resistance to a 5.0 rheostat by wire having a total resistance of 0.85-ohm?

total voltage = 4.5V, total resistance = 3.5 ohms, loop current = 4.5V / 3.5 ohms = 1.286Atotal voltage = 9V, total resistance = 4 ohms, loop current = 9V / 4 ohms = 2.25Atotal voltage = 13.5V, total resistance = 4.5 ohms, loop current = 13.5V / 4.5 ohms = 3Aetc.There is no solution to your problem conditions.


Is astm f2413-11 better than astm f2413-05?

ASTM F2413-11 is an updated version of ASTM F2413-05, incorporating advancements in safety and performance standards for protective footwear. The 2011 version includes improved test methods and criteria for impact and compression resistance, among other enhancements. Therefore, ASTM F2413-11 is generally considered better as it reflects the latest industry practices and safety requirements.


How multimeter work?

A multimeter is used to measure three electrical characteristics in circuits: Voltage Current and Resistance. AC and DC voltage and current can be measured. These three are related by Ohms law: Voltage ( in volts) = Current (in amps) times Resistance (in ohms) see http://en.wikipedia.org/wiki/Ohm%27s_law The main component in a multimeter is a micro amp meter. A hair thin wire is wound around an iron core on a pivot and with in a permanent magnetic field. When current flows through the coil it produces a magnetic force proportional to the current that causes the coil to move on the pivot against the permanent magnetic field. A pointer is attached to the coil and it move past the meter scale. The coil has some resistance 50 Ohms. If one volt of direct current is applied the 200 Milli amps of that current would turn the coil with so much force that it would pin the needle or break the mechanism. Typical micro ammeters are limited to 50 micro amps maximum. In a parallel circuit current divides inversely proportionally to the resistance. If a wire of 0.5 ohms is placed in parallel with the meter coil then 99% of the current will go through the wire (shunt) and 1% will go through the coil. The meter range would become 5000 micro amps or 5 Milli amps. If the shunt was made to .05 ohms the range would be 50 milli amps. Instead of a shunt a resistor of 4,950 ohms is wired in series with the coil. Now the total resistance is 5,000 ohms if 100 volts is placed across the meter and shunt the meter will only show 50 micro amps.


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A cell of emf 3.0 volts and internal resistance of 0.01 ohm is connected through an ammeter of 0.05-ohm internal resistance to a 5.0 rheostat by wire having a total resistance of 0.85-ohm?

total voltage = 4.5V, total resistance = 3.5 ohms, loop current = 4.5V / 3.5 ohms = 1.286Atotal voltage = 9V, total resistance = 4 ohms, loop current = 9V / 4 ohms = 2.25Atotal voltage = 13.5V, total resistance = 4.5 ohms, loop current = 13.5V / 4.5 ohms = 3Aetc.There is no solution to your problem conditions.


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