- Download PDF
Based on the picture, how would I figure out the capacitance Cu when R1=3200 ohm, R2=10 kohms, and C1=0.5uF? Assuming the bridge is balanced.
1 Answer | Add Yours
The figure is below.
The equilibrium condition for the bridge is `V_(CD) = 0 "Volts"` .
On the branch ACB of the bridge we have
`I_(ACB) = U/(R1+X_(C1) +R3)`
`V_(AC) = I_ACB*R1 = U* (R1)/(R1 +X_(C1) + R3)`
On the branch ADB of the bridge we have
`I_(ADB) = U/(R2 +X_(Cu))`
`V_(AD) =I(ADB)*R2 = U*(R2)/(R2+X_(Cu))`
As said above for equilibrium `V_(CD) = 0 "Volts"` which means
`V_(AC) = V_(AD)`
`(R1)/(R1+X_(C1) +R3) = (R2)/(R2+X_(Cu))`
`(R1)/(R3+ X_(C1)) = (R2)/X_(Cu)`
The above equation leads to
`(R1)/(j*omega*Cu) - (R2)/(j*omega*C1) = R2*R3`
Since the lest term is pure imaginary and right term is real, first condition that needs to be met for bridge equilibrium is R3=0.
Now we can write the second condition for bridge equilibrium as
`(R1)/(Cu) = (R2)/(C1)`
`Cu = C1* (R2)/(R1) =0.5*10000/3200 =1.5625 microF`
In practice the resistance R3 in series with C1 is provided to compensate for resistive looses in the unknown capacitor Cu. Therefore to measure Cu, first we modify the value of R3 for minimum indication of Voltmeter V, then we modify the value of R1 until the indication of voltmeter is zero.
We’ve answered 327,522 questions. We can answer yours, too.Ask a question