3 resistance each of 2 of which are equal are connected in series then the effective resistance is 30ohm when these resistance connected in parallel effective resistance is 3ohm find the individual resistance?
Asked by farzul hazan C.S | 13th Jul, 2013, 11:27: AM
Let the three resistances be R1, R 2 , and R3.
Let the effective Resistance in series be Rs, and in parallel be Rp.
Given: Rs = 30 ohm and Rp =3 ohm.
We know that Rs = R1 + R2 +R3 = 30
. Equation (1)
Equation (2)
If two of the three resistances are equal, then R1 =R2 =R (say).
From (1);
R+R+R3 =30
R3 = 30 2R
Equation (3)
From (2);
3(2R3+R) = RR3
6R3 +3R =RR3
Using (3); 6(30-2R) + 3R =R (30-2R)
180-12R+3R =30R-2R2
2R2 39R +180=0
Solving the quadratic equation,
Taking one possible value, we get
R1=12 , R2 =12 and R3=6 since R3=30-2R
Taking the other possible value, we get
R1=7.5 , R2=7.5 and R3 = 15 since R3=30-2R
Let the three resistances be R1, R 2 , and R3.
Let the effective Resistance in series be Rs, and in parallel be Rp.
Given: Rs = 30 ohm and Rp =3 ohm.
We know that Rs = R1 + R2 +R3 = 30 . Equation (1)
Equation (2)
If two of the three resistances are equal, then R1 =R2 =R (say).
From (1);
R+R+R3 =30
R3 = 30 2R Equation (3)
From (2);
3(2R3+R) = RR3
6R3 +3R =RR3
Using (3); 6(30-2R) + 3R =R (30-2R)
180-12R+3R =30R-2R2
2R2 39R +180=0
Solving the quadratic equation,
Taking one possible value, we get
R1=12 , R2 =12 and R3=6 since R3=30-2R
Taking the other possible value, we get
R1=7.5 , R2=7.5 and R3 = 15 since R3=30-2R
Answered by | 13th Jul, 2013, 08:43: PM
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