17. Four resistors are placed in series in a circuit Their resistances are R_(1)=10.1Omega ,R_(2)=19.8Omega R_(3)=50.3Omega ,R_(4)=25.4Omega . What is the total resistance of the circuit? a. 95.5Omega b. 108Omega C. 80.2Omega d. 106Omega

The total resistance of resistors in series is simply the sum of their individual resistances.<br /><br />* R_total = R1 + R2 + R3 + R4<br />* R_total = 10.1 Ω + 19.8 Ω + 50.3 Ω + 25.4 Ω<br />* R_total = 105.6 Ω<br /><br />Since 105.6 Ω is closest to 106 Ω, the correct answer is **d. 106 Ω**. The slight difference is likely due to rounding in the provided answer choices.<br />