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Question

A circuit has a \( 16 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three
resistors \( 2 \, \Omega \), \( 4 \, \Omega \), \( 8 \, \Omega \) in parallel. What is the current
through the \( 2 \, \Omega \) resistor?

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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. 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 = 2 + 1.14 = 3.14 Ω . Total current: I = (ε/Rtₒtₐl) = (16/3.14) ≈ 5.10 A . Voltage across

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