A battery of emf ε0=12 V is connected across a 4 m long uniform wire having resistance 4 Ω / m The cells of small emfs ε1=2 V and ε2=4 V having internal resistance 2 Ω and 6 Ω, respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point N, the distance of point N from the point A is equal to

Question

A battery of emf ε0=12 V is connected across a 4 m long uniform wire having resistance 4 Ω / m The cells of small emfs ε1=2 V and ε2=4 V having internal resistance 2 Ω and 6 Ω, respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point N, the distance of point N from the point A is equal to (A) (1/6) m (B) (1/3) m (C) 25 cm (D) 50 cm

Tardigrade Admin · Accepted Answer

Potential gradient, x=((12/8+16)) × 4=2 Vm -1 Effective emf of E1 and E2 E=((E1/r1)-(E2/r2)/1 / r1+1 / r2)=(1/2) text volt Balancing length =((1/2))((1/2))=(1/4) m =25 cm

$\frac{1}{6} m$

$\frac{1}{3} m$

25 cm

50 cm

Potential gradient,
$x = (\frac{12}{8 + 16}) \times 4 = 2 V m^{- 1}$
Effective emf of $E_{1}$ and $E_{2}$
$E = \frac{\frac{E _{1}}{r _{1}} - \frac{E _{2}}{r _{2}}}{1/ r _{1} + 1/ r _{2}} = \frac{1}{2} volt$
Balancing length $= (\frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} m = 25 c m$

61​m

31​m

25 cm

50 cm

Potential gradient, x=(8+1612​)×4=2Vm−1 Effective emf of E1​ and E2​ E=1/r1​+1/r2​r1​E1​​−r2​E2​​​=21​ volt Balancing length =(21​)(21​)=41​m=25cm

$\frac{1}{6} m$

$\frac{1}{3} m$

Potential gradient,
$x = (\frac{12}{8 + 16}) \times 4 = 2 V m^{- 1}$
Effective emf of $E_{1}$ and $E_{2}$
$E = \frac{\frac{E _{1}}{r _{1}} - \frac{E _{2}}{r _{2}}}{1/ r _{1} + 1/ r _{2}} = \frac{1}{2} volt$
Balancing length $= (\frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} m = 25 c m$