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A beginning RC circuit that resembles a wheatstone bridge

  1. Jun 13, 2009 #1
    Everybody knows what a wheatstone bridge looks like. R1 and R2 are on top and R3 and R4 are on bottom and R5 is in the middle of the angled box shape.

    let's replace R1 and R2 with 2 different capacitors. Let's also replace R5 with a switch.

    How would you figure out the potential between R3 and R4????
  2. jcsd
  3. Jun 13, 2009 #2


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    Assuming you have an AC supply, you work out the voltage across each capacitor by considering each C/R series circuit and then you subtract the voltages vectorially.

    If one of the capacitors was variable and known (and R3 = R4 ), you could adjust it until the bridge was balanced (using an AC detector like an oscilloscope) and then the two capacitors would be equal and you would know the unknown one. You would then have a capacitance bridge.
  4. Jun 13, 2009 #3
    the capacitors are NOT variable nor are they equal. The resistors are NOT variable nor are they equal. There is no amp meter nor is there a voltmeter in the middle...there is only a switch that has been closed. Now, how would you figure out the potential between R3 and R4. Also, the current is DC from a Battery.
    Last edited: Jun 13, 2009
  5. Jun 14, 2009 #4


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    To work out the voltage across across a capacitor in series with a resistor you draw a right angled triangle with , voltage across the resistor as the horizontal side and voltage across the capacitor along the vertical side, (usually downwards for capacitors) and total voltage across the hypotenuse.

    Assume 100 volts across the series combination.
    Take some random values for the capacitors and resistors.
    C1 300 ohms reactance
    C2 450 ohms reactance
    R3 270 ohms
    R4 120 ohms

    First, the left series combination:
    Impedance = sqrt ( 300 *300 + 270 * 270) =403.6 ohms
    Current = 100 / 403.6 = 0.247 A
    Voltage across resistor = 270 * 0.247 = 66.9 V
    Voltage across capacitor = 300 * 0.247 = 74.1 Volts
    Angle = tan-1(74.1 / 66.9) =-47.88 deg

    Next, the right series combination:
    Impedance = sqrt ( 450 *450 + 120 * 120) =465.7 ohms
    Current = 100 / 465.7 = 0.215 A
    Voltage across resistor =120 * 0.215 = 25.8 V
    Voltage across capacitor = 450 * 0.215= 96.75 Volts
    Angle = tan-1(96.75 / 25.8) = -75.06 deg

    You can't subtract these voltages because they are at different phase angles.
    Current in R3 is 0.247 A at +47.88 degrees relative to applied voltage.
    Voltage across R3 is (270 * .247) at +47.88 degrees =66.69 V at +47.75 deg relative to applied voltage
    Current in R4 is 0.215 A at 75.06 degrees relative to applied voltage
    Voltage across R4 is (120 * 0.215) or 11.46 volts at 75.06 degrees relative to applied voltage. Graphing these, I get a voltage of "about 60 volts".

    Solving it with trig, I get 56.66 volts

    All of this was with the switch open and AC. (did it say that before?)

    With the switch closed and DC, it is a lot easier.

    The capacitors behave like open circuits at DC so the capacitors are not conducting so there is no current in the resistors, so the voltage across the resistors is zero.
    Last edited: Jun 14, 2009
  6. Jun 14, 2009 #5
    You have to find the differential equation of the voltage as a function of time. Use laplace transforms of the capacitance value to find impedance ie 1/cs ohms, then use nodal and mesh analysis to find the function. Inverse transform them back into the time domain then you have the voltage function with time. In this case it would be easier to find the differential function for current. Use product over sum to merge the capacitance value to one value. also add up the other resistances. This becomes a simple diff eq if it is a DC source or an AC source. If it is a DC source the current approaches zero as a function of time, therefore at steady state all of the current will travel through the resistors.
    Last edited: Jun 14, 2009
  7. Jun 14, 2009 #6


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    With the switch closed and a DC supply, this just becomes two parallel resistors charging two parallel capacitors.

    After the charging is finished (depending on the time constant), all the supply voltage will be across the two capacitors in parallel.
  8. Jun 16, 2009 #7
    thank you for your help people!!! GREAT IDEAS!!!
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