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In the circuit, voltameter is ideal and its least count is 0.1 V (Whats least count?) The ammeter is 1 mA. Let reading of voltameter be 30.0 V and the reading of ammeter is 0.020 A. Calculate the value of resistance R with error limits. In the circuit, voltameter is ideal and its least count is 0.1 V (Whats least count?) The ammeter is 1 mA. Let reading of voltameter be 30.0 V and the reading of ammeter is 0.020 A. Calculate the value of resistance R with error limits.
In the circuit, voltameter is ideal and its least count is 0.1 V (Whats least count?) The ammeter is 1 mA. Let reading of voltameter be 30.0 V and the reading of ammeter is 0.020 A. Calculate the value of resistance R with error limits.
The least count(L.C) of any measuring equipment is the smallest quantity that can be measured accurately using that instrument.Thus Least Count indicates the degree of accuracy of measurement that can be achieved by the measuring instrument. Given, L.C of Ammeter is 1mA L.C of Voltameter is 0.1V Always an Ammeter is connected in-series- with circuit to measure electri current through it and Voltmeter is connected in parallel to measure voltage across its ends. OHMs law expression is : V=IR (or) R=V / I Here given I = 0.02A , V = 30V ,dV = 0.1V , dI = 10-3A so, R = 30 / 0.02 =1500 ohms ln R = ln V + ln I dR/R = dV/V + dI/I dR = R*(dV/V + dI/I ) =1500*(0.1/30 + 0.001/0.02) = 80 ohms Resistence with error limits = ( 1500 ± 80 ) ohms and % error in resistance is = 5.33% --- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
The least count(L.C) of any measuring equipment is the smallest quantity that can be measured accurately using that instrument.Thus Least Count indicates the degree of accuracy of measurement that can be achieved by the measuring instrument.
Given, L.C of Ammeter is 1mA
L.C of Voltameter is 0.1V
Always an Ammeter is connected in-series- with circuit to measure electri current through it and
Voltmeter is connected in parallel to measure voltage across its ends.
OHMs law expression is : V=IR (or) R=V / I
Here given I = 0.02A , V = 30V ,dV = 0.1V , dI = 10-3A
so, R = 30 / 0.02 =1500 ohms
ln R = ln V + ln I
dR/R = dV/V + dI/I
dR = R*(dV/V + dI/I )
=1500*(0.1/30 + 0.001/0.02)
= 80 ohms
Resistence with error limits = ( 1500 ± 80 ) ohms
and % error in resistance is = 5.33%
---
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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