A meter bridge with resistance R1 and R2 connect in two gap is balance...
**Introduction**
A meter bridge is a device used to measure an unknown resistance by comparing it with a known resistance. The balance point on the meter bridge indicates that the ratio of the two resistances is equal to the ratio of the distances from the zero end.
**Given Information**
- The meter bridge is balanced at 0.4 m from the zero end.
- The resistances in the two gaps are labeled as R1 and R2.
- A small resistance is connected in series with a 10 ohm resistor.
- The balance point shifts to 0.4 m from the other end.
- We need to find the value of the smaller resistance.
**Calculating the Ratio of Resistances**
The balance point on the meter bridge indicates that the ratio of the two resistances is equal to the ratio of the distances from the zero end. Let's calculate the ratio of resistances using the given information.
Ratio of resistances = (Distance from zero end to balance point) / (Total length of the meter bridge - Distance from zero end to balance point)
= 0.4 / (1 - 0.4)
= 0.4 / 0.6
= 2/3
**Calculating the New Balance Point**
When a small resistance is connected in series with the 10 ohm resistor, the effective resistance in that gap changes. Let's calculate the new balance point using the new effective resistance.
Ratio of resistances = (Distance from zero end to new balance point) / (Total length of the meter bridge - Distance from zero end to new balance point)
= 0.4 / (1 - 0.4)
= 0.4 / 0.6
= 2/3
From the given information, we know that the new balance point is 0.4 m from the other end. Let's assume this distance as x.
2/3 = x / (1 - x)
Solving this equation, we find that x = 2/5.
**Calculating the Value of the Smaller Resistance**
Now that we know the new balance point, we can calculate the value of the smaller resistance.
Ratio of resistances = (Resistance in the gap with the small resistance) / (Resistance in the gap with the 10 ohm resistor)
= x / (1 - x)
Substituting the value of x, we get:
Ratio of resistances = (2/5) / (1 - 2/5)
= (2/5) / (3/5)
= 2/3
From the given information, we know that the resistance in the gap with the 10 ohm resistor is 10 ohms. Let's assume the resistance in the gap with the small resistance as R.
2/3 = R / 10
Solving this equation, we find that R = 20/3 or approximately 6.67 ohms.
Therefore, the value of the smaller resistance is approximately 6.67 ohms.
A meter bridge with resistance R1 and R2 connect in two gap is balance...
R1/R2 = 0.4/0.6 = 4/6 = 2/3
this gives us
3/2*R1= R2_____(1)
Now according to given condition,
R1 + 10/ R2 = 0.6/0.4 = 6/4
R1 + 10 = 3/2R2____(2)
Put value of R2 from (1) in (2) we get,
R1 + 10 = 3/2 x 3/2*R1
R1 + 10 = 9/4R1
10 = 9/4R1 - R1
10 = 5/4 R1
R1 = 8ohms
R2 = 12 ohms
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