Practice question
Question
A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and three
resistors \( 2 \, \Omega \), \( 4 \, \Omega \), \( 8 \, \Omega \) in parallel. What is the current
through the \( 4 \, \Omega \) resistor?
Explanation
**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Parallel resistance: (1/R_p) = (1/2) + (1/4) + (1/8) = (4 + 2 + 1/8) = (7/8) ⇒ R_p = (8/7) ≈ 1.14 Ω . Total resistance: Rtₒtₐl = 1 + 1.14 = 2.14 Ω . Total current: I = (ε/Rtₒtₐl) = (10/2.14) ≈ 4.67 A . Voltage across parallel: V = I R_p = 4.67 × 1.14 ≈ 5.33 V
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