Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 6)
6.
A 1.2 k
resistor is in series with a 15 mH coil across a 10 kHz ac source. The magnitude of the total impedance is
resistor is in series with a 15 mH coil across a 10 kHz ac source. The magnitude of the total impedance is152.6
1,526
1,200
942
Discussion:
11 comments Page 1 of 2.
Mahesh said:
7 years ago
I think the right answer is 1200 Ω.
(2)
Srin said:
4 years ago
Good explanation, Thanks @Siddharth.
(1)
Debashri said:
1 decade ago
No the right ans is 1200 ohm.
Formula is xc=1/2 x 3.14 x 10kohm x 15mH
after that Z = root(R2+Xc2) than ans is 1200 ohm.
Formula is xc=1/2 x 3.14 x 10kohm x 15mH
after that Z = root(R2+Xc2) than ans is 1200 ohm.
Abhishek maji said:
1 decade ago
Yes, option B is correct answer.
Total impedance is given by, Z = square root of (R^2+XL^2).
Where, R=1.2*10^3 and XL=2*pi*10*10^3*15*10^-3.
That gives Z=1525.86 ohm.
Total impedance is given by, Z = square root of (R^2+XL^2).
Where, R=1.2*10^3 and XL=2*pi*10*10^3*15*10^-3.
That gives Z=1525.86 ohm.
Jignesh said:
1 decade ago
Simple when we connect 1.2 k resistor is in series with a 15 mH its must > 1.2k ohm so answer is definitively B.
Waqas habib said:
1 decade ago
Impedence of inductor is given by,
zl = 2*pi*f*l.
= 2*pi*10000*15/1000.
So answer B is correct.
zl = 2*pi*f*l.
= 2*pi*10000*15/1000.
So answer B is correct.
Madhu said:
1 decade ago
Option B correct.
R = 1200 ohms.
XL = (1/(2*pi*f*L)).
XL = (1/(2*3.14*10k*15m)).
Z = square root of(R2+XL2).
Z = 1526 ohms.
R = 1200 ohms.
XL = (1/(2*pi*f*L)).
XL = (1/(2*3.14*10k*15m)).
Z = square root of(R2+XL2).
Z = 1526 ohms.
Vishu said:
8 years ago
I didn't get exactly how to calculate impedance?
Can anyone explain it briefly?
Can anyone explain it briefly?
Oisin said:
8 years ago
Here, Z = √( R-squared + XL squared - XC squared).
ARKA said:
8 years ago
I think the right answer is 1200 ohm.
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