Electrical Engineering - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 4)
4.
A certain series circuit consists of a 1/8 W resistor, a 1/4 W resistor, and a 1/2 W resistor. The total resistance is 1200 
. If each resistor is operating in the circuit at its maximum power dissipation, total current flow is
. If each resistor is operating in the circuit at its maximum power dissipation, total current flow isAnswer: Option
Explanation:
Total P = P1 + P2 + P3 = 7/8w.
P = I2R.
I = sqrt (P/R)
Now, P = 7/8w, R = 1200 ohm.
Therefore,
I= sqrt ( (7/8)/1200 ) = 0.0270030862 = 27 mA.
Discussion:
32 comments Page 1 of 4.
Parashakthi p r said:
1 decade ago
P=I2R
Katta said:
1 decade ago
Nice explanation.
Sousheelya said:
1 decade ago
Appropriate explanation.
Raju said:
1 decade ago
Its good.
Raut Mahesh said:
1 decade ago
P=P1+P2+P3
P=(1/8)+(1/2)+(1/4)
P=(1+2+4)/8
P=7/8 W
& we have
P=I^2*R
Now P=7/8 w & R is given i.e. R=1200 ohm
Therefore,
I=sqrt(P/R)
I=sqrt((7/8)/1200)
I=27 mA
P=(1/8)+(1/2)+(1/4)
P=(1+2+4)/8
P=7/8 W
& we have
P=I^2*R
Now P=7/8 w & R is given i.e. R=1200 ohm
Therefore,
I=sqrt(P/R)
I=sqrt((7/8)/1200)
I=27 mA
Sibananda chakra said:
1 decade ago
Thank you nice explanation.
Jay chhalaliya said:
1 decade ago
But in parallel circuit formula of total power.
1/P = 1/P1+1/P2+1/P3.
1/P = 1/P1+1/P2+1/P3.
Dilip said:
1 decade ago
p = p1+p2+p3 = (1/8+1/2+1/4) = 7/8.
And, p = i^2*r,
And r = 1200,
i = sqrt(p/r).
Sqrt(7/8)/1200
i = 27mA.
And, p = i^2*r,
And r = 1200,
i = sqrt(p/r).
Sqrt(7/8)/1200
i = 27mA.
Thanushree said:
1 decade ago
I can't understood this problem please explain.
SELVARAJ said:
1 decade ago
Super and use full explain.
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