Question

How can the three resistances 2 Ω, 3 Ω, and 6 Ω be connected, so that (i) total resistance of 4 Ω is obtained, and (ii) total resistance of 1 Ω is obtained?

Hint:

Use the formula for series and parallel resistances:
\({Series: } R = R_1 + R_2\),
\({Parallel: } \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \).

Answer 1

(i) To obtain 4 Ω, we arrange \( 3Ω \) and \( 6Ω \) in parallel: \[ \frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{1}{2} \] So, \( R_p = 2Ω \). Adding \( 2Ω \) in series gives \( 4Ω \). (ii) To obtain 1 Ω, we connect all resistances in parallel: \[ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \] Thus, \( R = 1Ω \).