Find the equivalent resistance of the circuit given in figure
1 below (a) between the following points (i) A and B (ii) C and D (iii) E and F
(iv) A and F and (v) A and C. Numbers represent resistances in ohm.
1 below (a) between the following points (i) A and B (ii) C and D (iii) E and F
(iv) A and F and (v) A and C. Numbers represent resistances in ohm.
Solution
(i)
Resistance
between A and B
Resistance
between A and B
In this case, the entire circuit to the right side of AB is in
parallel with the 1Ω resistance connected directly across points A and B.
parallel with the 1Ω resistance connected directly across points A and B.
As seen, there are two parallel paths across points C and D;
one having a resistance of 6Ω and the other of (4+2) = 6Ω. The circuit becomes
as shown in figure 2 below.
one having a resistance of 6Ω and the other of (4+2) = 6Ω. The circuit becomes
as shown in figure 2 below.
From 2 above, the combined resistance, between C
and D is = 6||6 = 3Ω. The circuit is now reduced to the form shown in figure 3
below.
and D is = 6||6 = 3Ω. The circuit is now reduced to the form shown in figure 3
below.
From
figure 3 above, the 2Ω and 3Ω resistance are in series so they can be replaced
by a single (3+2) = 5Ω resistance. Then the circuit can now be further
simplified as shown in figure 4 below.
figure 3 above, the 2Ω and 3Ω resistance are in series so they can be replaced
by a single (3+2) = 5Ω resistance. Then the circuit can now be further
simplified as shown in figure 4 below.
From figure 4 above, the 1Ω and the 5Ω resistors are in
parallel. Therefore, the equivalent resistance = 1||5 =5/6 = 0.83Ω
parallel. Therefore, the equivalent resistance = 1||5 =5/6 = 0.83Ω
(ii)Resistance
Between C and D
As
seen from figure 1 above, there are three parallel paths between C and D. They
are CD of 6Ω, CEFD of (4+2) = 6Ω and CABD of (2+1) = 3Ω. Therefore, the
equivalent resistance is = 3||6||6 = 1.5Ω.
seen from figure 1 above, there are three parallel paths between C and D. They
are CD of 6Ω, CEFD of (4+2) = 6Ω and CABD of (2+1) = 3Ω. Therefore, the
equivalent resistance is = 3||6||6 = 1.5Ω.
(iii) Resistance
between E and F
between E and F
In this case, the circuit to the left side of EF is in
parallel with the 2Ω resistance connected directly across point E and F. This
left hand side circuit consist 4Ω resistance connected in series with a
parallel circuit of 6||(2+1) = 2Ω. Therefore the resistance to the left hand
side is 4+2 = 6Ω. Then the resistance between E and F is 6||2 = 1.5Ω.
parallel with the 2Ω resistance connected directly across point E and F. This
left hand side circuit consist 4Ω resistance connected in series with a
parallel circuit of 6||(2+1) = 2Ω. Therefore the resistance to the left hand
side is 4+2 = 6Ω. Then the resistance between E and F is 6||2 = 1.5Ω.
(iv)Resistance
between A and F
As we go from A to F, there are two
possible routes to begin with: one along ABDF and the other along AC. At point
C, there are again two alternatives, one along CDF and the other along CEF. The
circuit can be simplified to that shown in figure 5 shown below.
possible routes to begin with: one along ABDF and the other along AC. At point
C, there are again two alternatives, one along CDF and the other along CEF. The
circuit can be simplified to that shown in figure 5 shown below.
As seen from
figure 5 above, the equivalent resistance at CDEF is 6||6 = 3Ω. Therefore
equivalent resistance across A and F is 1||(2+3) = 5/6 = 0.83Ω.
figure 5 above, the equivalent resistance at CDEF is 6||6 = 3Ω. Therefore
equivalent resistance across A and F is 1||(2+3) = 5/6 = 0.83Ω.
(ii)
Resistance between A and C
Resistance between A and C
In this case, there are two parallel paths between A and C;
one is directly from A to C and the other along ABD. At point D, there are
again two parallel paths to C; one is directly along DC and the other is along
DFEC. As seen from above, the equivalent resistance at CDEF is 3Ω. Therefore,
equivalent resistance across A and C is = 2||(3+1) =8/6 =1.33Ω.
one is directly from A to C and the other along ABD. At point D, there are
again two parallel paths to C; one is directly along DC and the other is along
DFEC. As seen from above, the equivalent resistance at CDEF is 3Ω. Therefore,
equivalent resistance across A and C is = 2||(3+1) =8/6 =1.33Ω.
