answer me fast teacher
Solution
1. I is inversly propotional to R
as as R increases I that is current decreases.
2. V = IR
R = V/I = 24 V/ 120 mA = 0.2 x 103 Ohm
So
V = IR = 160 mA x 0.2 x 103 Ohm = 160x 10-3 x 0.2 x 103 = 32 V
3
L.C.=No. of divisions/ 0.5
Least count of voltmeter ie L.C = 20/ 0.5 =0 .025
4. V = IR
Nor R'= 4R at same V
V = I' R'
IR = I'R'
IR = I' 4R
I' = i/4
So new current is one fourth of original current
1. I is inversly propotional to R
as as R increases I that is current decreases.
2. V = IR
R = V/I = 24 V/ 120 mA = 0.2 x 103 Ohm
So
V = IR = 160 mA x 0.2 x 103 Ohm = 160x 10-3 x 0.2 x 103 = 32 V
3
L.C.=No. of divisions/ 0.5
Least count of voltmeter ie L.C = 20/ 0.5 =0 .025
4. V = IR
Nor R'= 4R at same V
V = I' R'
IR = I'R'
IR = I' 4R
I' = i/4
So new current is one fourth of original current