Practice question
Question
A \( 14 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 2 \,
\text{A} \) to a resistor. What is the resistance of the resistor?
Explanation
**EMF ε** is work done by non-electrostatic forces per unit charge, terminal voltage V = ε - I r, r internal resistance (Ω), I current (A). When external R = r, total resistance 2r, current I = ε/2r, power in external R is I²R = ε²/4r, total power ε²/2r, so half power dissipated externally, half internally. Terminal voltage: V = ε - I r = 14 - 2 × 2 = 10 V . Resistance: R = (V/I) = (10/2) = 5 Ω . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V =
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