Question

(i) In fig. find the potential difference between point A and B.
(ii) Now we wish to measure this potential difference by using a voltmeter of resistance $2k\Omega$. Find the reding of the voltmeter and percentage error.
(iii) solve part (ii) if the voltmeter were of resistance $20k\Omega$.
What conclusion do you draw from the results you get in the above parts?

Answer 1

(i) Apply voltage division rule then voltage across

$AB=40\times \frac{200}{400}$

$v_{AB}=10\times 2=20v$

(ii) In second cave, apply the given condition

again, apply voltage division rule $=20\times \frac{2000}{2200}$

voltmeter $v_{AB}=18.18v$

$\%$ error $=\frac{20-18.18}{20}=\frac{18.18-20}{20}$

$=-9.09\%$

Means $9.09\%$ reduces

i) If voltmeter resistance increased to $20k\Omega$ then,

again, apply voltage division rule $=20\times \frac{20000}{20,200}$

$v_{AB}=19.80v$.

$\%$ error $=\frac{19.80-20}{20}=-0.01$

means the voltagae reduced $1\%$ of actual value