Practice free →
HomeGATE EEelectricalengineeringElectric Circuits › Three resistors of $10\Omega$, $20\Omega$, $30\O…

Three resistors of $10\Omega$, $20\Omega$, $30\Omega$ are connected in SERIES. The total resistance is

A$5.45\Omega$
B$10\Omega$
C$60\Omega$
D$600\Omega$
Answer & Solution
Correct answer: C. $60\Omega$
1. SERIES: $R_{total} = R_1 + R_2 + R_3$. 2. $R_{total} = 10 + 20 + 30 = 60\Omega$. 3. Series resistance ALWAYS exceeds any individual resistor. 4. Option A is the parallel formula. Option B is just $R_1$. Option D multiplies instead of adds. _Source: Tony Kuphaldt, "Lessons in Electric Circuits — DC", Vol I, Ch 5 (Series resistors)._
Solve this in the app — GATE EE practice & 24k+ MCQs →
Related questions