Q.13
Q.14
Q.16
Q.18
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13.There are two possible combinations series and parallel
14.R=4
V=IR
I=V/R
I=3/4
14.R=4
V=IR
I=V/R
I=3/4
shrinivassk:
can u explain it long and step by step
Answered by
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13. When connected in series
3+3+3=9ohm
When connected in parallel
1/3+1/3+1/3=1/r
R= 1 ohm
When one connected in series with two parallel
3+3/2(the parallel)=9/2ohm
When two in a series and connected with one in parallel
6/3=2 ohm
14. Current drawn by the battery
R= 10ohm
V=3v
Then
I=V/R
I=3/10
18. The lenght of the nichrome wire will be 300cm
As the resistance is directly proportional to the length of the wire
R=rho L/A
16.let the resistance be R
Now as the wire is cut into three pieces and as we know the resistance is directly proportional to the length of the wire
New resistance of each wire=R/3
And when connected in parallel and calculated then
R=R/9ohm
And resistivity
Resistivity = RA/L
Now the area A remains same
Resistance is =R/9
And lenght L = L/3
New resistivity = resistivity/3
The second part of question 16
The total current will remain same I.
But current in each wire will be I/3
3+3+3=9ohm
When connected in parallel
1/3+1/3+1/3=1/r
R= 1 ohm
When one connected in series with two parallel
3+3/2(the parallel)=9/2ohm
When two in a series and connected with one in parallel
6/3=2 ohm
14. Current drawn by the battery
R= 10ohm
V=3v
Then
I=V/R
I=3/10
18. The lenght of the nichrome wire will be 300cm
As the resistance is directly proportional to the length of the wire
R=rho L/A
16.let the resistance be R
Now as the wire is cut into three pieces and as we know the resistance is directly proportional to the length of the wire
New resistance of each wire=R/3
And when connected in parallel and calculated then
R=R/9ohm
And resistivity
Resistivity = RA/L
Now the area A remains same
Resistance is =R/9
And lenght L = L/3
New resistivity = resistivity/3
The second part of question 16
The total current will remain same I.
But current in each wire will be I/3
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