Problem #3, the bridge http://arts-sciences.und.edu/physics-astrophysics/gpl/_fi

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Problem #3, the bridge
http://arts-sciences.und.edu/physics-astrophysics/gpl/_files/docs/lab-105-w-bridge.pdf

easy problem but havent learned

Assume the negative terminal of the battery (short line) is at zero volts, even though there is no ground symbol in the circuit. Let the battery voltage be Vo. (a) Assuming that no current flows through Rv, how is the current that flows through R1 related to the current that flows through R2 ? Notice that current flows through the two in series, in this case. (b) Use the previous conclusion to figure out the voltage at a, assuming no current in Rv. Show the algebra, not just the result.   Va = (c) Still assuming the bridge is balanced (meaning what ? ) find the voltage at point b. Thus Vb =    (d) If the bridge is balanced, what is the relation between Vb and Va ? (e) Thus using the results you have gotten, prove that when the bridge is balanced, (R2/R1)=(Rx/Rs)


Problem #3, the bridge
http://arts-sciences.und.edu/physics-astrophysics/gpl/_files/docs/lab-105-w-bridge.pdf

easy problem but havent learned

Assume the negative terminal of the battery (short line) is at zero volts, even though there is no ground symbol in the circuit. Let the battery voltage be Vo. (a) Assuming that no current flows through Rv, how is the current that flows through R1 related to the current that flows through R2 ? Notice that current flows through the two in series, in this case. (a) Assuming that no current flows through Rv, how is the current that flows through R1 related to the current that flows through R2 ? Notice that current flows through the two in series, in this case. (b) Use the previous conclusion to figure out the voltage at a, assuming no current in Rv. Show the algebra, not just the result.   Va = (c) Still assuming the bridge is balanced (meaning what ? ) find the voltage at point b. Thus Vb =    (d) If the bridge is balanced, what is the relation between Vb and Va ? (e) Thus using the results you have gotten, prove that when the bridge is balanced, (R2/R1)=(Rx/Rs)
(b) Use the previous conclusion to figure out the voltage at a, assuming no current in Rv. Show the algebra, not just the result.   Va = (c) Still assuming the bridge is balanced (meaning what ? ) find the voltage at point b. Thus Vb =    (d) If the bridge is balanced, what is the relation between Vb and Va ? (e) Thus using the results you have gotten, prove that when the bridge is balanced, (R2/R1)=(Rx/Rs)

Explanation / Answer

In the given setup,when there is no flow in current through the resistor Rv,then the setup is said to be a balanced bridge circuit.

thus ,by ohms law

Vb-Va=I*Rv=0*Rv=0

=>Vb=Va ---(1)

a)

when no current is flowing through Rv,

at point A,

incoming current =outgoing current

=>I through R1=I through R2

b)

along R1,v and R2

as,I through R1=I through R2

=> (Va-0)/R1=(V-Va)/R2

=>Va/R1=V/R2-Va/R2

=>Va/R1+Va/R2=V/R2

=>Va(1/R1+1/R2)=V/R2

=>Va[(R1+R2)/R1*R2]=V/R2

=>Va[(R1+R2)/R1]=V

=>Va=V[R1/(R1+R2)] ---(2)

c)from ohm's law ,we got equation --(1)

=>Vb=Va

d)

as,I through Rx=I through Rs

(Vb-0)/Rx=(V-Vb)/Rs

=>Vb/Rx=V/Rs-Vb/Rs

=>Vb/Rx+Vb/Rs=V/Rs

=>Vb(1/Rx+1/Rs)=V/Rs

=>Vb[(Rx+Rs)/Rx*Rs]=V/Rs

=>Vb[(Rx+Rs)/Rx]=V

=>Vb=V[Rx/(Rx+Rs)] ---(3)

as Vb=Va

=> from 2 and 3

V[Rx/(Rx+Rs)] =V[R1/(R1+R2)]

=>[Rx/(Rx+Rs)]=[R1/(R1+R2)]

=>[(Rx+Rs)/Rx]=[(R1+R2)/R1] [subtracting 1 both the side]

=>Rs/Rx =R2/R1

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