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Zorluk: OrtaElectric Circuits and Measuring Instruments

A cell supplies a current of 0.6 A0.6\text{ A} when connected across a 2.0 Ω2.0\text{ }\Omega resistor. When the resistor is replaced with a 5.0 Ω5.0\text{ }\Omega resistor, the current supplied by the cell drops to 0.3 A0.3\text{ A}. What is the internal resistance of the cell?

  1. 1.0 Ω1.0\text{ }\OmegaCevap
  2. B
    0.0 Ω0.0\text{ }\Omega
  3. C
    3.0 Ω3.0\text{ }\Omega
  4. D
    0.5 Ω0.5\text{ }\Omega

Cevap

The internal resistance of the cell is 1.0 Ω1.0\text{ }\Omega.
The electromotive force (e.m.f.) EE of a cell is given by E=I(R+r)E = I(R + r), where II is the current, RR is external resistance, and rr is internal resistance. Setting up equations for both cases: E=0.6(2.0+r)E = 0.6(2.0 + r) and E=0.3(5.0+r)E = 0.3(5.0 + r). Equating them gives 1.2+0.6r=1.5+0.3r1.2 + 0.6r = 1.5 + 0.3r, which simplifies to 0.3r=0.30.3r = 0.3, yielding r=1.0 Ωr = 1.0\text{ }\Omega.

Adım Adım Çözüm

1
Formulate the electromotive force equation for the first circuit setup
E=I1(R1+r)=0.6(2.0+r)=1.2+0.6rE = I_1(R_1 + r) = 0.6(2.0 + r) = 1.2 + 0.6r
The total voltage supplied by the cell equals the total current multiplied by the sum of external and internal resistance.
2
Formulate the electromotive force equation for the second circuit setup
E=I2(R2+r)=0.3(5.0+r)=1.5+0.3rE = I_2(R_2 + r) = 0.3(5.0 + r) = 1.5 + 0.3r
The cell's electromotive force EE and internal resistance rr remain unchanged despite changing the external resistor.
3
Equate both expressions for electromotive force and solve for internal resistance
1.2+0.6r=1.5+0.3r  ⟹  0.3r=0.3  ⟹  r=1.0 Ω1.2 + 0.6r = 1.5 + 0.3r \implies 0.3r = 0.3 \implies r = 1.0\text{ }\Omega
Equating the two expressions allows direct solution for the single unknown variable rr.

Anahtar Kavram

Electromotive Force and Internal Resistance of Cells
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Electric Circuits and Measuring Instruments

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