Question

In the circuit given, AB is a uniform wire of resistance 15 ohms and length 1m. It is connected to a cell E1 of emf 2V and negligible internal resistance and a resistance R. The balancing point with another cell E2 of emf 75 mV is found at 30 cm from end A. Calculae the value of the resistance R.

Answer 1

Hope it helps..

E1 =2V

E2=75*10^-3V

L1=30cm

Lab=1m

Rab=15 ohm

Answer 2

Answer:

The value of the resistance R is 105 ohm.

Explanation:

Given that,

Resistance $R_{AB} = 15\ ohm$

Length l = 1 m=100 cm

emf E₁ = 2 V

emf E₂ = 75 m V

length l₂ =30 cm

Current down from cell,

$I = \dfrac{E_{1}}{R_{AB}+R}$

$I= \dfrac{2}{15+R}$...(I)

Potential drop across wire

$V_{AB}=I\times R$

Put the value of I from equation (I)

$V_{AB}=\dfrac{2}{15+R}\times15$

Now, potential gradient along the wire

$x =\dfrac{V_{AB}}{100}$....(II)

Here, x= potential gradient

$V_{AB}$= potential drop

put the value of $V_{AB}$ in equation (II)

$x=\dfrac{30}{100(15+R)}$

At balance point, the emf

$E_{2}=kl_{2}$

$75\times10^{-3}=\dfrac{30}{100(15+R)}\times30$

$15+R=\dfrac{9000}{75}$

$R= 120-15=105\ Omega$

Hence, The value of the resistance R is 105 ohm.