Example 2 (Worked Example on Equivalent Resistant).

Two resistors of values 1kΩ and
4kΩ are connected in series across a constant voltage supply of 100V. A
voltmeter having an internal resistance of 12kΩ is connected across the 4kΩ
resistor. Draw the circuit and calculate
(a)   
True voltage across 4kΩ resistor before the voltmeter was
connected.
(b)   
Actual voltage across 4kΩ resistor after the voltmeter was
connected and the voltage recorded by the voltmeter
(c)   
Change in supply current when voltmeter is connected.
(d)   
Percentage error in voltage across 4kΩ resistor.
Solution

(a)   
True voltage drop across 4kΩ resistor before the voltmeter was
connected is found by voltage-divider rule
=100 x (4/5) = 80V.
(b)   
In figure 1 above, voltmeter has been joined across the 4kΩ
resistor. The equivalent resistance between B and C

 

= 3kΩ

(a)   
Change is supply current when voltmeter is connected.
Equivalent resistance without voltmeter = 1+4 =5kΩ.

Equivalent
resistance with voltmeter = 1 + (4||12)
= 1 + 3 = 4kΩ

 












The reduction in the value of voltage being
measured is called voltmeter loading effect because voltmeter loads down the
circuit element across which it is connected. Smaller the voltmeter resistance
as compared to the resistance across which it is connected, greater the loading
effect and hence, greater the error in the voltage reading. Loading effect
cannot be avoided but can be minimized by selecting a voltmeter of resistance
much higher than that of the network across which it is connected.

0 0 votes
Article Rating
Subscribe
Notify of
guest

0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x