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iampaul
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Sorry for this post, but I am desperate for an answer. Please reply thanks in advance
Consider a simple circuit composed of a battery whose ends are connected by a wire. The potential difference of the battery is Vbatt
ohm's law gives: V=IR
If the resistance of the circuit is ignored then the potential drop across the wire is zero.
But using kirchhoffs voltage law the sum of the potential difference in the loop must be zero,my first question is where does the potential drop?
If the resistance of the wire is not ignored then there is a voltage drop equal to the potential difference of the battery.Even when the resistance is small approaching zero the voltage drop is still equal to Vbatt. The limit approached by the voltage drop as the resistance approaches zero is still Vbatt but the voltage drop when the resistance equals zero is zero by using V=IR. My 2nd question is why is it not continuous? Shouldn't the voltage drop still be Vbatt
My third question is about a switch in parallel with a resistor and a battery. The resistance of the conducting wires is ignored. My professor said that if the switch is open then there is zero current passing through it and using V=IR the voltage across the switch is zero. Because it is in parallel with the resistor the potential difference across the resistor is also zero. but can't a switch be considered as a conducting material(air) with high resistance. The total resistance of the parallel combination of the switch and the resistor is approximately equal to just the resistance of the resistor by using the equation: (1/rs)+(1/r)=(1/rT) since the ratio (1/rs) almost vanishes because rs, the resistance of the switch is too big. But since there is a total resistance there must be a voltage across the resistor.
Please reply
Consider a simple circuit composed of a battery whose ends are connected by a wire. The potential difference of the battery is Vbatt
ohm's law gives: V=IR
If the resistance of the circuit is ignored then the potential drop across the wire is zero.
But using kirchhoffs voltage law the sum of the potential difference in the loop must be zero,my first question is where does the potential drop?
If the resistance of the wire is not ignored then there is a voltage drop equal to the potential difference of the battery.Even when the resistance is small approaching zero the voltage drop is still equal to Vbatt. The limit approached by the voltage drop as the resistance approaches zero is still Vbatt but the voltage drop when the resistance equals zero is zero by using V=IR. My 2nd question is why is it not continuous? Shouldn't the voltage drop still be Vbatt
My third question is about a switch in parallel with a resistor and a battery. The resistance of the conducting wires is ignored. My professor said that if the switch is open then there is zero current passing through it and using V=IR the voltage across the switch is zero. Because it is in parallel with the resistor the potential difference across the resistor is also zero. but can't a switch be considered as a conducting material(air) with high resistance. The total resistance of the parallel combination of the switch and the resistor is approximately equal to just the resistance of the resistor by using the equation: (1/rs)+(1/r)=(1/rT) since the ratio (1/rs) almost vanishes because rs, the resistance of the switch is too big. But since there is a total resistance there must be a voltage across the resistor.
Please reply
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