Electrical measuring instruments

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In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is $l = 25 cm$ . If ‘R’ is replaced by a wire of half length and same diameter that of R of same material. Then the balancing distance $l (in cm)$ will now be

Moderate

Solution

$I n i t i a l l y, \frac{S}{R} = \frac{75}{25} =3 N o w, R= \frac{ρ l}{A} = \frac{4 ρ l}{{πd}^{2}} ∴ R^{'} = \frac{4 ρ (l / 2)}{{πd}^{2}} = \frac{R}{2} F i n a l l y, \frac{S}{R'} = (\frac{100 - l}{l}) \Rightarrow \frac{S}{(R/2)} = (\frac{100 - l}{l}) \Rightarrow \frac{2S}{R} = (\frac{100 - l}{l}) \Rightarrow 2 \times 3 = \frac{100 - l}{l} \Rightarrow l = \frac{100}{7} = 14.29 cm$

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Electrical measuring instruments

Remember concepts with our Masterclasses.

In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is l = 25 cm . If ‘R’ is replaced by a wire of half length and same diameter that of R of same material. Then the balancing distance l in cm will now be

Initially, SR=7525=3 Now, R=ρlA=4ρlπd2 ∴ R'=4ρl/2πd2=R2 Finally, SR'=100−ll ⇒SR/2=100−ll ⇒2SR=100−ll ⇒2×3=100−ll ⇒l=1007=14.29 cm

Ready to Test Your Skills?

free study material

In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is $l = 25 cm$ . If ‘R’ is replaced by a wire of half length and same diameter that of R of same material. Then the balancing distance $l (in cm)$ will now be