In a parallel circuit, the total resistance is always less than the smallest individual resistor.

In a parallel circuit, the total resistance is always less than the smallest individual resistor.

• A. Sometimes
• B. Never
• C. False
• D. True

Answer: (D)

In a parallel arrangement, multiple paths for current open up, so the overall resistance drops. The math behind this is 1/R_total = 1/R1 + 1/R2 + ...; since each term is positive, adding them makes the sum larger than any individual 1/Ri, which means R_total must be smaller than each individual resistance. For example, 4 Ω and 6 Ω in parallel give R_total = 1/(1/4 + 1/6) = 2.4 Ω, which is less than both 4 Ω and 6 Ω. So with multiple resistors in parallel, the total resistance is always less than the smallest resistor. If there were only one resistor, the total would equal that resistor, but with multiple parallel paths the rule holds.

In a parallel arrangement, multiple paths for current open up, so the overall resistance drops. The math behind this is 1/R_total = 1/R1 + 1/R2 + ...; since each term is positive, adding them makes the sum larger than any individual 1/Ri, which means R_total must be smaller than each individual resistance. For example, 4 Ω and 6 Ω in parallel give R_total = 1/(1/4 + 1/6) = 2.4 Ω, which is less than both 4 Ω and 6 Ω. So with multiple resistors in parallel, the total resistance is always less than the smallest resistor. If there were only one resistor, the total would equal that resistor, but with multiple parallel paths the rule holds.