Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 22)
22.
The parallel combination of a 470 
resistor and a 1.2 k resistor is in series with the parallel combination of three 3 k resistors. A 200 V source is connected across the circuit. The resistor with the most current has a value of
resistor and a 1.2 k resistor is in series with the parallel combination of three 3 k resistors. A 200 V source is connected across the circuit. The resistor with the most current has a value of470 or 1.2 k
or 1.2 k3 k
470
1.2 k
Discussion:
10 comments Page 1 of 1.
Chaman chauhan said:
8 years ago
@vignesh and Bhanu
As per question 1.2 k ohm and 470 ohm resistor are in series with three 3 k ohm resistor is correct but before that they are parallel with each other so the current will divide in both 1.2 k ohm and 470 ohm resistor according to their value and we know that lower the resistor higher the current value so the 470 ohm is correct answer.
As per question 1.2 k ohm and 470 ohm resistor are in series with three 3 k ohm resistor is correct but before that they are parallel with each other so the current will divide in both 1.2 k ohm and 470 ohm resistor according to their value and we know that lower the resistor higher the current value so the 470 ohm is correct answer.
Arun kumar M said:
8 years ago
Least R = 400 &ohms;. So the current is high.
Bhanu said:
8 years ago
Yes, I agree @Ashwath.
Ashwath said:
10 years ago
I agree with @Vignesh.
Current through 1.2 Kohm & 470 ohm are same.
So the answer is 1.2 Kohm & 470 ohm which is not provided in the option.
Current through 1.2 Kohm & 470 ohm are same.
So the answer is 1.2 Kohm & 470 ohm which is not provided in the option.
Vignesh said:
10 years ago
Answer is wrong because in series circuit, the current will be same. As per the question 1.2K ohm and 470 ohm are in series with parallel of 3k ohm resistor.
(1)
Saeed niazi said:
1 decade ago
= (470*1200)/(470+1200)+(3k*3k*3k/9k).
= 337.7+27k/9k.
= 337.7+3.
= 340.
V=IR =>I=V/R => I=0.587 throughout
Now current divider.
Across 470,
I = (1200/1200+470)*0.587 = 0422.
Across 1200(1.2k),
I = (470/470+1200)*0.587 = 0.281.
And in each 3k current will be 0.587/3 = 0.19.
So the most current is through 470 ohm.
= 337.7+27k/9k.
= 337.7+3.
= 340.
V=IR =>I=V/R => I=0.587 throughout
Now current divider.
Across 470,
I = (1200/1200+470)*0.587 = 0422.
Across 1200(1.2k),
I = (470/470+1200)*0.587 = 0.281.
And in each 3k current will be 0.587/3 = 0.19.
So the most current is through 470 ohm.
(2)
Ahmed said:
1 decade ago
In calculation way how prove that current is low in high resistor or high in low resistor.
First find the total resistor.
Rt= 1337.72.
Then use ohm's law to obtain the source current use that current in Current divider rule or obtain how much current in each loop resistors.
First find the total resistor.
Rt= 1337.72.
Then use ohm's law to obtain the source current use that current in Current divider rule or obtain how much current in each loop resistors.
(1)
Jagdish Prasad said:
1 decade ago
But we know that current is always maximum in low resistor.
Pradeep said:
1 decade ago
Thanks maansa.
Manasa said:
1 decade ago
As current takes a least resistive path. Hence current in 470 ohms will be the maximum.
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