please solve this with explanation
Attachments:
Answers
Answered by
1
you see top form traingle ,
both above side of traingle in series
so , Req of these two resistance is (2 + 2)ohm = 4 ohm
now , Req join with base of traingle in parellel so , R"eq of these = 4 x 2 /(4+2) ohm = 4/3 ohm
now ,
R" eq , and rest resistance join in series
so final equivalent resistance
= 2 + 4/3 + 2 = 16/3 ohm
so , 16/3 ohm is answer
both above side of traingle in series
so , Req of these two resistance is (2 + 2)ohm = 4 ohm
now , Req join with base of traingle in parellel so , R"eq of these = 4 x 2 /(4+2) ohm = 4/3 ohm
now ,
R" eq , and rest resistance join in series
so final equivalent resistance
= 2 + 4/3 + 2 = 16/3 ohm
so , 16/3 ohm is answer
shaili1:
i have not understood
Answered by
0
First of all, you need to find the resistance between the points B and C,
Now, the points B,A,C are connected is series so, they can be replaced by an equivalent resistance Rs.
Rs = 2+2 = 4 Ω
The points B,A,C and B,C are connected in parallel so, they can be replaced by an equivalent resistance Rp.
1/Rp = 1/4+1/2
1/Rp = 3/4 Ω
Rp = 4/3 Ω
The total resistance between the points B and C = 3/4 Ω
Now, the points D,B,C,E and B,C are connected in series so, they can be replaced by an equivalent resistance Rs.
Rs = 2+4/3+2 = 16/3 Ω
∴ The resistance between the points D and E = 16/3 Ω
Now, the points B,A,C are connected is series so, they can be replaced by an equivalent resistance Rs.
Rs = 2+2 = 4 Ω
The points B,A,C and B,C are connected in parallel so, they can be replaced by an equivalent resistance Rp.
1/Rp = 1/4+1/2
1/Rp = 3/4 Ω
Rp = 4/3 Ω
The total resistance between the points B and C = 3/4 Ω
Now, the points D,B,C,E and B,C are connected in series so, they can be replaced by an equivalent resistance Rs.
Rs = 2+4/3+2 = 16/3 Ω
∴ The resistance between the points D and E = 16/3 Ω
Similar questions