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Zorluk: ZorElectric Circuits and Measuring Instruments

A potentiometer wire of length 100 cm100\text{ cm} has a resistance of 10 Ω10\text{ }\Omega. It is connected in series with a driver cell of electromotive force 3.0 V3.0\text{ V} and internal resistance 2.0 Ω2.0\text{ }\Omega, alongside an external series resistor of 8.0 Ω8.0\text{ }\Omega. A test cell of unknown electromotive force EE gives a balance point at a length of 60 cm60\text{ cm} from the zero end of the wire. What is the value of EE in volts?

Cevap: 0.9 V

Cevap

The electromotive force of the test cell is 0.9 V0.9\text{ V}.
The e.m.f. of the test cell is balanced by the potential difference across a length of 60 cm60\text{ cm} of the potentiometer wire. Accounting for the driver cell's internal resistance (2.0 Ω2.0\text{ }\Omega), external series resistor (8.0 Ω8.0\text{ }\Omega), and wire resistance (10 Ω10\text{ }\Omega), the total resistance of the primary circuit is 20.0 Ω20.0\text{ }\Omega. This yields a primary current of 0.15 A0.15\text{ A} and a potential drop across the wire of 1.5 V1.5\text{ V}. The resulting potential gradient is 0.015 V/cm0.015\text{ V/cm}, which when multiplied by the balance length of 60 cm60\text{ cm} gives an e.m.f. of 0.9 V0.9\text{ V}.

Adım Adım Çözüm

1
Calculate total resistance in the primary driver circuit
Rtotal=10 Ω+2.0 Ω+8.0 Ω=20.0 ΩR_{total} = 10\text{ }\Omega + 2.0\text{ }\Omega + 8.0\text{ }\Omega = 20.0\text{ }\Omega
The driver cell's internal resistance, potentiometer wire, and external series resistor are connected in series.
2
Calculate the current flowing through the potentiometer wire
I=3.0 V20.0 Ω=0.15 AI = \frac{3.0\text{ V}}{20.0\text{ }\Omega} = 0.15\text{ A}
Apply Ohm's law to the complete primary circuit.
3
Find the voltage drop across the potentiometer wire
Vwire=0.15 A×10 Ω=1.5 VV_{wire} = 0.15\text{ A} \times 10\text{ }\Omega = 1.5\text{ V}
The potential difference across the wire depends on its resistance and the primary current.
4
Determine the potential gradient along the wire
k=1.5 V100 cm=0.015 V/cmk = \frac{1.5\text{ V}}{100\text{ cm}} = 0.015\text{ V/cm}
Potential gradient is the potential drop per unit length of the wire.
5
Calculate the e.m.f. of the unknown test cell
E=0.015 V/cm×60 cm=0.9 VE = 0.015\text{ V/cm} \times 60\text{ cm} = 0.9\text{ V}
At the balance point, no current flows from the test cell, so its e.m.f. equals the potential drop across the balance length.

Anahtar Kavram

Potentiometer principle and potential gradient
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