Question 1.38: Resistors of 22 Ω, 47 Ω, and 33 Ω are connected (a) in serie......

(a) In the series circuit R=R_{1}+R_{2}+R_{3}, thus R=22  \Omega+47  \Omega+33  \Omega=102  \Omega

(b) In the parallel circuit:

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

Thus,

\frac{1}{R}=\frac{1}{22  \Omega}+\frac{1}{47  \Omega}+\frac{1}{33  \Omega}

Or,

\frac{1}{R}=0.045+0.021+0.03

from which,

\frac{1}{R}=0.096=10.42  \Omega