SATHEE: Current Electricity 4 Question 11

Current Electricity 4 Question 11

11. On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10 c m$ . The resistance of their series combination is $1 k Ω$ . How much was the resistance on the left slot before interchanging the resistances?

(2018 Main)

(a) $990 Ω$

(b) $910 Ω$

(d) $550 Ω$

Show Answer

Solution:

$\frac{R_{1}}{R_{2}} = \frac{l}{100 - l}$

and $\frac{R_{2}}{R_{1}} = \frac{l - 10}{100 - (l - 10)}$

From Eqs. (i) and (ii),

$\Rightarrow l = 55 c m$

Substituting in Eq. (i), we get

$\frac{R_{1}}{R_{2}} = \frac{55}{45}$

Further,

$R_{1} + R_{2} = 100 Ω$

Solving Eqs. (iii) and (iv),

$R_{1} = 550 Ω$

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Current Electricity 4 Question 11

11. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10cm. The resistance of their series combination is 1kΩ. How much was the resistance on the left slot before interchanging the resistances?

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11. On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10 c m$ . The resistance of their series combination is $1 k Ω$ . How much was the resistance on the left slot before interchanging the resistances?