Electrical Engineering - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 21)
21.
A series circuit consists of three resistors. Two resistors are 1.2 k
each. The total resistance is 12 k. The value of the third resistor
each. The total resistance is 12 k. The value of the third resistoris 9,600
is 960
is 1.2 k
cannot be determined
Discussion:
6 comments Page 1 of 1.
Raaz said:
1 decade ago
R1+R2+R3 = 12-(1) Than we know the value of R1 and R2 because two resistor value is 1.2 each, So R1=1.2 or even R2=1.2 so next,
1.2+1.2+R3 = 12 from equation (1), then
2.4+R3 = 12 s
R3 = 12-2.4 = 9.6 kilo ohm so,
9.6*1000=96/10*1000=9600 because 1kilo=1000.
1.2+1.2+R3 = 12 from equation (1), then
2.4+R3 = 12 s
R3 = 12-2.4 = 9.6 kilo ohm so,
9.6*1000=96/10*1000=9600 because 1kilo=1000.
(1)
Geetha said:
1 decade ago
R1 = 1.2*10^3.
R2 = 1.2*10^3.
R = 12*10^3.
R3 = ?
R = R1+R2+R3.
12*10^3 = ((1.2+1.2)*10^3) + R3.
12*10^3 = (2.4*10^3) + R3.
R3 = (12-2.4)*10^3.
R3 = 9600 ohm.
R2 = 1.2*10^3.
R = 12*10^3.
R3 = ?
R = R1+R2+R3.
12*10^3 = ((1.2+1.2)*10^3) + R3.
12*10^3 = (2.4*10^3) + R3.
R3 = (12-2.4)*10^3.
R3 = 9600 ohm.
Raviteja said:
1 decade ago
Let unknown R3 = x;
R1+R2+R3 = Req;
1.2+1.2+x = 12;
x = 12-2.4 = 9.6 kohm.
R1+R2+R3 = Req;
1.2+1.2+x = 12;
x = 12-2.4 = 9.6 kohm.
(1)
Ravi said:
1 decade ago
r1+r2=2.4
total resistance is 12kohm
12-2.4=9.6
9.6*100=9600
total resistance is 12kohm
12-2.4=9.6
9.6*100=9600
Yellapulovakishore said:
6 years ago
1.2+1.2+X=2.4,
X=24/2.4.
X=9.6kilo ohms,
X=9600Ω.
X=24/2.4.
X=9.6kilo ohms,
X=9600Ω.
Kiran said:
1 decade ago
r1+r2+r3=R
1.2+1.2+X=12
x=9.6 Kiloohm
9600 ohm
1.2+1.2+X=12
x=9.6 Kiloohm
9600 ohm
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