Given four 1 _ resistors, state how they must be connected to give an overall resistance of (a) 1 Ω (b) 1Ω (c) 11 Ω (d) 2 1 Ω, all four resistors being connected
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
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