A battery of emf E 0-12 V is connected across a 4 m long uniform wire having resistance 4 - m - The cells of small emfs E 1-2 V and E 2-4 V having internal resistance 2 - and 6 - 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 toA- 2-5 mB- 0-75 mC- 3 mD- 1-25 m

The correct option is D 1.25 m Equivalent emf of ε_1 and ε_2 is ε = ε_1/r_1 + ε_2/r_2/1/r_1 + 1/r_2 = 2.5 volt Now V_AN = ε ∴ (12/8+16)(4l) = 2.5 Solving we ge ...

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Byju's Answer

Standard XII

Physics

Parallel Combination of Cells

A battery of ...

Question

A battery of emf $E_{0} = 12 V$ is connected across a 4{m} long uniform wire having resistance $\frac{4 Ω}{m}$ . The cells of small emfs $E_{1} = 2 V a n d E_{2} = 4 V$ 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

2.5 m

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3 m

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1.25 m

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0.75 m

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Solution

The correct option is C
1.25 m

Equivalent emf of $ε_{1} a n d ε_{2}$ is
$ε = \frac{\frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}}} = 2.5 v o l t$
Now $V_{A N} = ε$
$∴ (\frac{12}{8 + 16}) (4 l) = 2.5$
Solving we get, l = 1.25 m

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The correct option is C 1.25 m Equivalent emf of ε1 and ε2 is ε=ε1r1+ε2r21r1+1r2=2.5volt Now VAN=ε ∴(128+16)(4l)=2.5 Solving we get, l = 1.25 m

The correct option is C
1.25 m

Equivalent emf of $ε_{1} a n d ε_{2}$ is
$ε = \frac{\frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}}} = 2.5 v o l t$
Now $V_{A N} = ε$
$∴ (\frac{12}{8 + 16}) (4 l) = 2.5$
Solving we get, l = 1.25 m