Series Circuit Physics Example 3

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Example 3

hard
A 6 V6 \text{ V} battery powers a series circuit with a 3 Ω3 \text{ } \Omega resistor and an unknown resistor. The current is 0.5 A0.5 \text{ A}. Find the unknown resistance and the power dissipated by each resistor.

Solution

  1. 1
    Total resistance: RT=VI=60.5=12 ΩR_T = \frac{V}{I} = \frac{6}{0.5} = 12 \text{ } \Omega.
  2. 2
    Unknown resistance: R2=123=9 ΩR_2 = 12 - 3 = 9 \text{ } \Omega.
  3. 3
    Power in R1R_1: P1=I2R1=(0.5)2×3=0.75 WP_1 = I^2 R_1 = (0.5)^2 \times 3 = 0.75 \text{ W}.
  4. 4
    Power in R2R_2: P2=I2R2=(0.5)2×9=2.25 WP_2 = I^2 R_2 = (0.5)^2 \times 9 = 2.25 \text{ W}.

Answer

R2=9 Ω;P1=0.75 W,P2=2.25 WR_2 = 9 \text{ } \Omega; \quad P_1 = 0.75 \text{ W}, P_2 = 2.25 \text{ W}
In a series circuit, the larger resistor dissipates more power since current is the same through both. Total power (3 W3 \text{ W}) equals VI=6×0.5VI = 6 \times 0.5.

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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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