Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is 40 cm. Now an unknown resistance $x$ is connected in series with Q and new balancing length is found to be 25 cm measured from the same end. Then the value of $x$ will be _____ $Ω$ .

Question

Smart Achievers · Accepted Answer

(6)

$\frac{4}{Q} = \frac{40}{60} \Rightarrow Q = 6$

$\frac{4}{6 + x} = \frac{25}{75} \Rightarrow x = 6$

Solution

Q.
Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is 40 cm. Now an unknown resistance $x$ is connected in series with Q and new balancing length is found to be 25 cm measured from the same end. Then the value of $x$ will be _____ $Ω$ .

Ans.
(6)

$\frac{4}{Q} = \frac{40}{60} \Rightarrow Q = 6$

$\frac{4}{6 + x} = \frac{25}{75} \Rightarrow x = 6$

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Solution

Q. Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length l1 is 40 cm. Now an unknown resistance x is connected in series with Q and new balancing length is found to be 25 cm measured from the same end. Then the value of x will be _____ Ω.

Ans. (6) 4Q=4060⇒Q=6 46+x=2575⇒x=6

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Q.
Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is 40 cm. Now an unknown resistance $x$ is connected in series with Q and new balancing length is found to be 25 cm measured from the same end. Then the value of $x$ will be _____ $Ω$ .

Ans.
(6)

$\frac{4}{Q} = \frac{40}{60} \Rightarrow Q = 6$

$\frac{4}{6 + x} = \frac{25}{75} \Rightarrow x = 6$