Soru

Zorluk: OrtaElectric Circuits and Measuring Instruments

A potentiometer wire of length 100 cm100\text{ cm} has a resistance of 5.0 Ω5.0\ \Omega. It is connected in series with a driver cell of e.m.f. 3.0 V3.0\text{ V} (having negligible internal resistance) and a protective series resistor RsR_s. A balance point is obtained at 60.0 cm60.0\text{ cm} along the wire for a cell of e.m.f. 1.2 V1.2\text{ V}. What is the value of the series resistor RsR_s in ohms?

Cevap: 2.5 \Omega

Cevap

The resistance of the series resistor RsR_s is 2.5 Ω2.5\ \Omega.
At balance, the potential drop across the 60.0 cm60.0\text{ cm} portion of the potentiometer wire equals the test cell e.m.f. (1.2 V1.2\text{ V}). Since the 100 cm100\text{ cm} wire has a total resistance of 5.0 Ω5.0\ \Omega, the 60.0 cm60.0\text{ cm} section has a resistance of 3.0 Ω3.0\ \Omega. This requires a main circuit current of I=1.2 V3.0 Ω=0.4 AI = \frac{1.2\text{ V}}{3.0\ \Omega} = 0.4\text{ A}. The total driver circuit resistance is Rtotal=3.0 V0.4 A=7.5 ΩR_{total} = \frac{3.0\text{ V}}{0.4\text{ A}} = 7.5\ \Omega. Subtracting the wire's 5.0 Ω5.0\ \Omega resistance gives Rs=2.5 ΩR_s = 2.5\ \Omega.

Adım Adım Çözüm

1
Calculate the resistance of the balanced portion of the potentiometer wire.
Rbalance=5.0 Ω×60.0 cm100 cm=3.0 ΩR_{balance} = 5.0\ \Omega \times \frac{60.0\text{ cm}}{100\text{ cm}} = 3.0\ \Omega
The resistance of a uniform wire is directly proportional to its length.
2
Determine the current in the main potentiometer circuit.
I=1.2 V3.0 Ω=0.4 AI = \frac{1.2\text{ V}}{3.0\ \Omega} = 0.4\text{ A}
At the balance point, no current flows through the galvanometer branch, so the potential difference across the balance length equals the e.m.f. of the test cell.
3
Compute the total resistance of the driver circuit.
Rtotal=3.0 V0.4 A=7.5 ΩR_{total} = \frac{3.0\text{ V}}{0.4\text{ A}} = 7.5\ \Omega
According to Ohm's law, total resistance equals the total e.m.f. divided by the circuit current.
4
Calculate the value of the series resistor RsR_s.
Rs=7.5 Ω5.0 Ω=2.5 ΩR_s = 7.5\ \Omega - 5.0\ \Omega = 2.5\ \Omega
The total resistance is the sum of the potentiometer wire resistance and the series resistance.

Anahtar Kavram

Potentiometer balance condition and circuit analysis
Tahmini Süre:1m 30s
Bu soruyu puanla