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.
Yuvaraj said:
6 years ago
The current is same for all the components in a series circuit and here, all three resistors have different power ratings which means only one of the three resistor is operating at its maximum power dissipation and others are consuming less or more amount of power compared their power rating.
Adil 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 & are 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 & are is given i.e. R = 1200 ohm.
Therefore,
I = sqrt (P/R).
I = sqrt ((7/8)/1200).
I = 27 mA.
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
Veena said:
9 years ago
Nice @Punith asked the good question because in competitive examination calculator is not allowed. So anyone will explain how to calculate square roots without calculator?
Prince said:
10 years ago
P = P1+P2+P3= 1/8+1/4+1/2= (1+2+4)/8;
P = 7/8.
We know that P = I^2R ; 7/8 = I^2x 1200;
Then, I = sqrt (7/8)/1200 = 0.027003 A = 27.003m A.
P = 7/8.
We know that P = I^2R ; 7/8 = I^2x 1200;
Then, I = sqrt (7/8)/1200 = 0.027003 A = 27.003m A.
Rohine said:
3 years ago
I2 =(7/8)*(1/1200).
= √ of 0.000729.
= √ of 729*10^-6 =sqrt root of 729*(10^-3)2.
=27*10^-3.
=27 milli amp.
= √ of 0.000729.
= √ of 729*10^-6 =sqrt root of 729*(10^-3)2.
=27*10^-3.
=27 milli amp.
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.
Rishika said:
9 years ago
I don't know how to calculate in calculator can anyone help us who are having trouble about this, please.
Mitra rai said:
3 years ago
*I2 =(7/8)*(1/1200).
= √ of 0.000729.
= √ of 729*10^-6 = √729*(10^-3)2.
=27*10^-3.
=27 milli amp.
= √ of 0.000729.
= √ of 729*10^-6 = √729*(10^-3)2.
=27*10^-3.
=27 milli amp.
PAVAN said:
9 years ago
Yes @Punith, same doubt for me.
How to calculate the square root value without a calculator?
How to calculate the square root value without a calculator?
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