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penroseandpaper

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- Homework Statement
- (b) The circuit shown in Figure 1 can be used to measure the resistance of a platinum resistance thermometer (PRT). AB is a uniform resistance wire of length 1.0 m and C is a sliding contact on this wire. A standard resistor R is included in the circuit. The position of C is adjusted until the voltmeter V reads zero.

(i) By applying Kirchhoff’s laws to loops ADCA and BCDB, deduce an expression for the resistance of the PRT in terms of l1, l and the value of the standard resistor.

(ii) The PRT consists of 9.0 m of wire of diameter 8.0 × 10−2 mm. The voltmeter reads 0 V when l1 = 0.44 m. If the standard resistor, R, has a resistance of 224 Ω, what is the resistivity of platinum?

- Relevant Equations
- Kirchoff's Laws

Hi,

I found this question/s online and I was wondering if somebody was able to explain the answer given, specifically the first bit that says R1/(R1+R2) = R4/(R3+R4) (is this something to do with an equivalent resistor to replace the two in series on each side? But why are R1 and R4 the numerators?)

Also, is it easy to convert the log value at the end with e in it to one with just 10 to the power of something?

Let R1 be the resistance of the left part of the wire, R2 the right part, R3 the "R" and R4 the PRT. L1 and L2 the lengths of the left and right part of the wire, L the total lengths.

Your use of "I" for the lengths is too confusing as I usually means current. Don't need loop equations.

The current is zero when the voltage at the junction of R1,2 is equal to that at the junction of R3,4.

Therefore it is balanced when R1/(R1+R2) = R4/(R3+R4)

in terms of the wire, length is proportional to the resistance, so

L1/(L1+L2) = R4/(R3+R4)

L1/(L) = R4/(R3+R4)

(R3+R4)L1 = R4L

R3L1+R4L1 = R4L

R4(L–L1) = R3L1

R4 = R3L1 / (L–L1)"The PRT consists of 9.0 m of wire of diameter 8.0 × 10−2 mm. The voltmeter reads 0 V when l1 = 0.44 m. If the standard resistor, R, has a resistance of 224 Ω, what is the resistivity of platinum?"

8.0 × 10−2 mm ??

guessing that that is 0.08 mm. Please use correct terminology in the future.

R4 = R3L1 / (L–L1) = 224•0.44 / (1–0.44) = 176 Ω

R = ρL/A

r = 0.04 mm = 4e-5 m

A = πr² = 5.03e-9 m²

ρ = AR/L = 5.03e-9•176/9 = 9.83e-8 Ω-mResistance of a wire in Ω

R = ρL/A

ρ is resistivity of the material in Ω-m

L is length in meters

A is cross-sectional area in m²

A = πr², r is radius of wire in m

resistivity Pt 105e-9 Ω-m

I found this question/s online and I was wondering if somebody was able to explain the answer given, specifically the first bit that says R1/(R1+R2) = R4/(R3+R4) (is this something to do with an equivalent resistor to replace the two in series on each side? But why are R1 and R4 the numerators?)

Also, is it easy to convert the log value at the end with e in it to one with just 10 to the power of something?

The answer given was:The answer given was:

Let R1 be the resistance of the left part of the wire, R2 the right part, R3 the "R" and R4 the PRT. L1 and L2 the lengths of the left and right part of the wire, L the total lengths.

Your use of "I" for the lengths is too confusing as I usually means current. Don't need loop equations.

The current is zero when the voltage at the junction of R1,2 is equal to that at the junction of R3,4.

Therefore it is balanced when R1/(R1+R2) = R4/(R3+R4)

in terms of the wire, length is proportional to the resistance, so

L1/(L1+L2) = R4/(R3+R4)

L1/(L) = R4/(R3+R4)

(R3+R4)L1 = R4L

R3L1+R4L1 = R4L

R4(L–L1) = R3L1

R4 = R3L1 / (L–L1)"The PRT consists of 9.0 m of wire of diameter 8.0 × 10−2 mm. The voltmeter reads 0 V when l1 = 0.44 m. If the standard resistor, R, has a resistance of 224 Ω, what is the resistivity of platinum?"

8.0 × 10−2 mm ??

guessing that that is 0.08 mm. Please use correct terminology in the future.

R4 = R3L1 / (L–L1) = 224•0.44 / (1–0.44) = 176 Ω

R = ρL/A

r = 0.04 mm = 4e-5 m

A = πr² = 5.03e-9 m²

ρ = AR/L = 5.03e-9•176/9 = 9.83e-8 Ω-mResistance of a wire in Ω

R = ρL/A

ρ is resistivity of the material in Ω-m

L is length in meters

A is cross-sectional area in m²

A = πr², r is radius of wire in m

resistivity Pt 105e-9 Ω-m