AB is a potentiometer wire of length 100 cm and its resistance is 10 - It is connected in series with a battery of emf 2 V and resistance 40 - If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire- the value of E isA- 0-8 VB- 1-6 VC- 0-08 VD- 0-16 V

The correct option is D 0.16 VPotential gradient nbsp; across potentiometer wire is nbsp; dV/dl = E_0 R_AB/(R_AB +R)L = 2 × 10/(10 + 40)1 = 0.4 Vm^ 1 U ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Electric Bulbs

AB is a poten...

Question

$A B$ is a potentiometer wire of length $100 c m$ and its resistance is $10 Ω$ . It is connected in series with a battery of emf $2 V$ and resistance $40 Ω$ . If a source of unknown emf $E$ is balanced by $40 c m$ length of the potentiometer wire, the value of $E$ is

0.8 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.6 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.08 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.16 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $0.16 V$
Potential gradient across potentiometer wire is

$\frac{d V}{d l} = \frac{E_{0} R_{A B}}{(R_{A B} + R) L}$
$= \frac{2 \times 10}{(10 + 40) 1}$
$= 0.4 V m^{- 1}$

Unknown emf is
$E = \frac{d V}{d x} \times l$ $(∵ l = 40 c m)$
$E = 0.4 \times 40 \times 10^{- 2} = 0.16 V$

Suggest Corrections

Similar questions

A B

is a potentiometer wire of length

100 cm

and its resistance is

10 Ω

. It is connected in series with a resistance

R = 40 Ω

and a battery of emf

2 V

and negligible internal resistance. If a source of unknown emf

E

is balanced by

40 cm

length of the potentiometer wire, then the value of

E

Q. A potentiometer wire of length

1 m

has a resistance of

100 Ω

. It is connected in series with a resistance and a battery of emf

2 V

of negligible resistance. A source of emf

10 m V

is balanced against a length of

40 c m

of the potentiometer wire. What is the value of the external resistance.

A potentiometer wire of length $100 c m$ has a resistance of $10 Ω$ . It is connected in series with an external resistance and a cell of emf $2 V$ and negligible internal resistance. A source of emf $10 m V$ is balanced against a length of $40 c m$ of the potentiometer wire.Then the value of external resistance is

A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of E is:

Q. AB ia a potentiate wire of length 100 cm and is resistance is

10 Ω

. It is connected in series with a resistance

R = 40 Ω

and a battery of emf 2 V and negligible internal resistance. If a source of unknown potentiate wire, the value of E is:

Join BYJU'S Learning Program

AB is a potentiometer wire of length 100 cm and its resistance is 10 Ω. It is connected in series with a battery of emf 2 V and resistance 40 Ω. If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire, the value of E is

The correct option is D 0.16 VPotential gradient across potentiometer wire is dVdl=E0RAB(RAB+R)L =2×10(10+40)1 =0.4 Vm−1 Unknown emf is E=dVdx×l (∵l=40 cm) E=0.4×40×10−2=0.16 V

$A B$ is a potentiometer wire of length $100 c m$ and its resistance is $10 Ω$ . It is connected in series with a battery of emf $2 V$ and resistance $40 Ω$ . If a source of unknown emf $E$ is balanced by $40 c m$ length of the potentiometer wire, the value of $E$ is