1. What is the potential difference DeltaVab between points a and also b as soon as the switch is open? Closed?

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2. DeltaV = IR3. I fixed I for the battery, which is 8A. What perform I usage for my R value to settle for DeltaV in between a and b, though? I tried, for instance, DeltaV = IR = 8A*6ohms = 48V. I guess I"m perplexed as to what I use for R to calculate the potential at each allude, which are the values I usage to get my answer. Could someone describe this to me conceptually? For the potential at A, for instance, I tried multiplying the present (8A) by 3ohms because the current flows via that 3ohm resistor to obtain to point A...
What intermediate measures did you use , to gain 8 Amp via the switch open up ?you considered R_left = 6 Ohm , and also R_best = 6 Ohm ...so : what"s the voltage throughout R_left ? Voltage throughout R_right?exactly how much of this 8 Amp goes wbelow ? (this charge is conserved, you know)V = I R is a statement about ONE tool .
The voltage throughout the bottom left resistor = the supply voltage times 3 split by (the amount of the two series left resistors) .So it is 24 * 3 / (3 + 3) volts.Similarly, The voltage across the bottom appropriate resistor = the supply voltage times 1 split by (the amount of the two series ideal resistors) .So, it is 24 * 1 / (5 +1 ) volts.Then, what is the voltage between the points a and also b because the bottoms of these resistors are joined?
The voltage throughout the bottom left resistor = the supply voltage times 3 divided by (the sum of the 2 series left resistors) .So it is 24 * 3 / (3 + 3) volts.Similarly, The voltage throughout the bottom right resistor = the supply voltage times 1 separated by (the amount of the 2 series ideal resistors) .So, it is 24 * 1 / (5 +1 ) volts.Then, what is the voltage in between the points a and b considering that the bottoms of these resistors are joined?
Are you saying the voltage difference is 12V - 4V = 8V? Also, why carry out we treatment about the bottom left and also ideal resistors especially, considering that the present flows through the peak 2 resistors to acquire to points a and b before getting to the bottom two resistors?
Are you saying the voltage distinction is 12V - 4V = 8V? Also, why execute we care about the bottom left and best resistors especially, since the present flows through the top 2 resistors to acquire to points a and b prior to getting to the bottom 2 resistors?
Yes, that is right.This is just two voltage dividers.In each voltage divider the voltage throughout each reduced resistor is equal to (the resistance of that resistor separated by the full resistance) times the supply voltage.You might say the current in the left two resistors is 24 V / 6 ohms = 4 amps, so the voltage throughout the reduced resistor is 4 amps * 3 ohms or 12 volts. (Ohm"s Law)Then the existing in the right side is 24 volts / 6 ohms = 4 amps. So, the voltage across the 1 ohm resistor is 4 amps * 1 ohm or 4 volts.So, the distinction in voltage at a - b is 8 volts. a is at +12 volts loved one to the negative supply and also b is at +4 volts relative to the negative supply.The first method is much faster, though.You can read about voltage dividers here :http://en.wikipedia.org/wiki/Voltage_divider
Mar 11, 2010#7
Linus Pauling
1900

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Thank you!Now, what is deltaV between points a and b once the switch is closed? In this instance, the peak 2 (3 and also 5 ohm) resistors are in parallel, as are the bottom 2 (3 and 1 ohm). The identical resistor for each of those pairs are in series through one one more, so the present need to be the exact same into and out of them. Solving I for the top 2 resistors:<(1/3)+(1/5)>-1=1.875.I = V/R = 24V / 1.875ohm = 12.8AThen, V for point a: V = IR = (12.8A)(3ohm) = 38.4Vfor allude b: V = (12.8A)(1ohm) = 12.8VdeltaV = 25.6 V.... is this correct?Also, in fixing for a and also b, I offered the resistence of the bottom two resistors. If I use the two, I obtain a deltaV of the very same magnitude however opposite sign (-25.6V). What does this mean?EDIT: transforms out 25.6V is incorrect....???