A potentiometer wire of length 1 m has a resistance of 10 ohm. if it is connected to 6 v battery qith resistance of 5 ohm
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HEY DEAR ...
Given ,
length of wire , l = 100cm
Resistance, R = 10Ω
emf of battery , E1 = 6V
Resistance , R1 = 5Ω
& x = 40 cm
Now , Current , I = E1/R+R1
⇒ I = 6/10+5 = 2/5 A
Also , V = IR
⇒V = 2/5 × 10 = 4V
Now , emf of primary cell = Vx/l
= 4×40/100 = 1.6V
HOPE , IT HELPS ...
Given ,
length of wire , l = 100cm
Resistance, R = 10Ω
emf of battery , E1 = 6V
Resistance , R1 = 5Ω
& x = 40 cm
Now , Current , I = E1/R+R1
⇒ I = 6/10+5 = 2/5 A
Also , V = IR
⇒V = 2/5 × 10 = 4V
Now , emf of primary cell = Vx/l
= 4×40/100 = 1.6V
HOPE , IT HELPS ...
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