Physics
Grade-9
Easy

Question

Two wires of the same material have the same length, but their radii are in the ratio of 5:3. They are combined in series, where the resistance of the thicker wire is 12 ohms. Calculate the total resistance of the combination

  1. 40 Ω
  2. 12 Ω
  3. 33.3 Ω
  4. 0.03 Ω

hintHint:

The total resistance of the combination of resistors in series is  RT= R1+R2

The correct answer is: 33.3 Ω


    When two wires of radii in the ratio of 5:3 are combined in series, where the resistance of the thicker wire is 12 ohms, the resistance of the combination is 32 Ω.
    • Given, the radii of the wire are in the ratio of 5:3 which is R2/R1
    • We can write it as, R2/R1= 5/3, where R2= (5/3) R1
    • The R1 given is the thicker wire with resistance of 12 Ω. So R2=5 over 3 cross times 12 equals 20capital omega
    • Hence, their total resistance is RT= R1+R2, i.e. 12+20= 32 Ω

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