Circuit Physics Example 4

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Example 4

medium
A circuit has a battery and three identical bulbs. Two bulbs are in parallel with each other, and this parallel pair is in series with the third bulb. If the battery is 6 V6 \text{ V} and each bulb has resistance 6 Ω6 \text{ } \Omega, what is the current from the battery?

Solution

  1. 1
    Parallel pair: Rpair=6×66+6=3612=3 ΩR_{\text{pair}} = \frac{6 \times 6}{6 + 6} = \frac{36}{12} = 3 \text{ } \Omega.
  2. 2
    Total resistance: RT=Rpair+R3=3+6=9 ΩR_T = R_{\text{pair}} + R_3 = 3 + 6 = 9 \text{ } \Omega.
  3. 3
    Current from battery: I=VRT=69=0.667 AI = \frac{V}{R_T} = \frac{6}{9} = 0.667 \text{ A}

Answer

I≈0.667 AI \approx 0.667 \text{ A}
Real circuits often combine series and parallel elements. Analyzing them requires systematically reducing parallel groups to equivalent resistances, then treating the result as a simple series circuit.

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