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Question

A \( 10 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance is connected to two resistors
\( 4 \, \Omega \) and \( 6 \, \Omega \) in parallel. What is the current through the \( 6 \, \Omega \)
resistor?

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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) = (1/4) + (1/6) = (3 + 2/12) = (5/12) ⇒ R_p = (12/5) = 2.4 Ω . Total resistance: Rtₒtₐl = 2 + 2.4 = 4.4 Ω . Total current: I = (10/4.4) ≈ 2.27 A . Voltage across parallel: V = I R_p = 2.27 × 2.4 ≈ 5.45 V . Current through 6 Ω :

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