Question

Three resistors of 1 Ω, 0.1 Ω and 0.01 Ω are connected in series combination. Their equivalent resistance is

• A. 1.11 Ω
• B. 1.2 Ω
• C. 2.01 Ω
• D. 2.1 Ω

Answer 1

The Correct Option is A

For resistors connected in series, the total or equivalent resistance \( R_{\text{eq}} \) is simply the sum of the individual resistances: \[ R_{\text{eq}} = R_1 + R_2 + R_3 \] Given:
\( R_1 = 1 \, \Omega \),
\( R_2 = 0.1 \, \Omega \),
\( R_3 = 0.01 \, \Omega \).
Substituting the values: \[ R_{\text{eq}} = 1 + 0.1 + 0.01 = 1.11 \, \Omega \] Thus, the equivalent resistance is \( 1.11 \, \Omega \).

The correct option is (A): \(1.11\ Ω\)