Electronics and Communication Engineering - Measurements and Instrumentation - Discussion
Discussion Forum : Measurements and Instrumentation - Section 1 (Q.No. 27)
27.
A meter with a resistance of 100 Ω and a full scale deflection of current of 1 mA is to be converted into voltmeter of 0 - 5 V range. The multiplier resistance should be
Answer: Option
Explanation:
.
Discussion:
5 comments Page 1 of 1.
Suyash Dixit said:
9 years ago
R total = V/ I
=> 5 V / 1mA
=> 5 K ohm
As we know that, [1 k ohm = 1000 ohm ]
So,
R multiplier = 5000 ohm - 100 ohm
=> 4900 ohm.
=> 5 V / 1mA
=> 5 K ohm
As we know that, [1 k ohm = 1000 ohm ]
So,
R multiplier = 5000 ohm - 100 ohm
=> 4900 ohm.
Prem ku parida said:
9 years ago
R = V - v/ig.
v = ig * Rg V = range voltage.
R = 5 - 1/1000 * 100 = 5 - .1 = 4.9/ig = 4.9 * 1000 = 4900ohm.
v = ig * Rg V = range voltage.
R = 5 - 1/1000 * 100 = 5 - .1 = 4.9/ig = 4.9 * 1000 = 4900ohm.
Chirag said:
1 decade ago
How is this calculated? Please explain.
Debashis panda said:
5 years ago
Yes it's right, thanks @Suyash Dixit.
Krish said:
1 decade ago
Rs = v/Im-Rm.
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