हिंदी

What is the End Error in a Meter Bridge? How is It Overcome? the Resistances in the Two Arms of the Metre Bridge Are R = ω and S Respectively. When the Resistance

Advertisements
Advertisements

प्रश्न

Answer the following question.
What is the end error in a meter bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = Ω and S respectively.  When the resistance S is shunted with equal resistance, the new balance length found to be 1.5 l1, where l2 is the initial balancing length. calculate the value of s.

संक्षेप में उत्तर
Advertisements

उत्तर

The shifting of zero of the scale at different points as well as the stray resistance gives rise to the end error in meter bridge wire. This error arises due to the non-uniformity of the meter wire End corrections can be estimated by including known resistances P1 and Q1 in the two ends and finding the null point.

We have

R = 5 Ω

According to wheat stone bridge principle:

`R/l_1 = S/(100 - l_1)`

`5/l_1 = S/(100 - l_1)`    ....(1)

After shunting, connect the resistance in parallel

`S → S/2` 

`5/(1.5 l_1) = S/(2(100 - 1.5 l_1))`  ....(2)

Thus, equation (1) can be written as 

`500 - 5l_1 = Sl_1`                       ....(3)

And equation 2 is 

`10(100 - 1.5l_1) = 1.5 Sl_1`       ....(4)

From equation (3) and (4)

`(500 - 5l_1)/l_1 = (1000 - 15l_1)/(1.5l_1)`

`750 - 7.5l_1 = 1000 - 15l_1`

`-250 = -7.5l_1`

`l_1 = 100/3`

S = `(500 - (5 xx 100)/3)/(100/3) = (500 - 500/3)/(100/3) = 1000/3 xx 3/100`

 S = 10 Ω

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2018-2019 (March) 55/1/3

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?


The equivalent resistance between points. a and f of the network shown in Figure 2 is :

a) 24 Ω

b) 110 Ω

c) 140 Ω

d) 200 Ω


Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?


Can the potential difference across a battery be greater than its emf?


The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0°C and 0.001°C. Find the thermo-emf (Seebeck emf) developed. For bismuth-silver, a = − 46 × 10−6 V°C−1 and b = −0.48 × 10−6 V°C−2.


An energy source will supply a constant current into the load if its internal resistance is ______.

The internal resistance of dry cell is ...A..., than the internal resistance of common electrolytic cell. Here, A refers to ______.

A cell having an emf E and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by ______.


A cell E1 of emf 6 V and internal resistance 2 Ω is connected with another cell E2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is ______.


A cell of emf E is connected across an external resistance R. When current 'I' is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance 'r' of the cell is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×