Given four 1_resistors, state how they must be connected to give an overall resisance of (a);a 12 () 1a 2, ll four resistors being connected in each case.
Answers
Answer:
(a) \textbf{\textup{All four in parallel}}All four in parallel (see Fig. 5.16), since
\frac{1}{R}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=\frac{4}{1}
R
1
=
1
1
+
1
1
+
1
1
+
1
1
=
1
4
i.e. R = \frac{1}{4}\, \OmegaR=
4
1
Ω
(b) \textbf{Two in series, in parallel with another two in series}Two in series, in parallel with another two in series (see Fig. 5.17), since 1\, \Omega1Ω and 1\, \Omega1Ω in series gives 2\, \Omega2Ω , and 2\, \Omega2Ω in parallel with 2\, \Omega2Ω gives
\frac{2\times 2}{2+2}=\frac{4}{4}=1\, \Omega
2+2
2×2
=
4
4
=1Ω
(c) \textbf{Three in parallel, in series with one}Three in parallel, in series with one (see Fig. 5.18), since for the three in parallel,
\frac{1}{R}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=\frac{3}{1}
R
1
=
1
1
+
1
1
+
1
1
=
1
3
i.e. R = \frac{1}{3}\, \OmegaR=
3
1
Ω and \frac{1}{3}\, \Omega
3
1
Ω in series with 1\, \Omega1Ω gives 1\, \Omega1Ω
(d) \textbf{Two in parallel, in series with two in series}Two in parallel, in series with two in series (see Fig. 5.19), since for the two in parallel
R=\frac{1\times 1}{1+1}=\frac{1}{2}\, \OmegaR=
1+1
1×1
=
2
1
Ω, and \frac{1}{2}\, \Omega, \, 1 \, \Omega
2
1
Ω,1Ω and 1\, \Omega1Ω in series gives 2\frac{1}{2}\, \Omega2
2
1
Ω