Series Circuit Physics Example 4

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Example 4

medium
Three resistors (4 Ω4 \text{ } \Omega, 6 Ω6 \text{ } \Omega, and 2 Ω2 \text{ } \Omega) are connected in series to a 24 V24 \text{ V} battery. Calculate: (a) total resistance, (b) current, (c) voltage across the 6 Ω6 \text{ } \Omega resistor.

Solution

  1. 1
    (a) In series: RT=4+6+2=12 ΩR_T = 4 + 6 + 2 = 12 \text{ } \Omega.
  2. 2
    (b) I=VRT=2412=2 AI = \frac{V}{R_T} = \frac{24}{12} = 2 \text{ A} (same through all components).
  3. 3
    (c) V6=IR6=2×6=12 VV_6 = IR_6 = 2 \times 6 = 12 \text{ V}.

Answer

(a)  12 Ω;(b)  2 A;(c)  12 V(a)\; 12 \text{ } \Omega; \quad (b)\; 2 \text{ A}; \quad (c)\; 12 \text{ V}
In a series circuit, resistances add directly, current is the same everywhere, and voltage divides proportionally across each resistor (VRV \propto R).

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A circuit arrangement in which components are connected end-to-end along a single path, so exactly the same current flows through every component.

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