Question 498: Current Electricity

Question

The voltmeter has range 10 V and its internal resistance is $50 \Omega$. To increase the range of voltmeter to 15 V , the resistance which is to be connected is

Accepted Answer

Step 1: Understand
By default, voltmeter is rated 10Ã¢ÂÂ¯V Ã¢ÂÂ At 10 V, the voltmeter draws current
$   I = \frac{V}{R} = \frac{10}{50} = 0.2 	ext{ A}   $
So the
full-scale deflection current
= 0.2 A.
Step 2: New Range
We want voltmeter to read up to 15 V (instead of 10 V).
For 15 V applied, same current (0.2 A) must flow through the meter.
More voltage Ã¢ÂÂ we must drop extra 5 V somewhere with some series resistance.
Step 3: Series Resistance Calculation.
At full-scale deflection, current through meter = 0.2 A (constant).
Additional resistance:
$ R_{series} = \frac{V_{extra}}{I} = \frac{5}{0.2} = 25 \ \Omega $
Step 4: Check Options
Option A: 125Ã¢ÂÂ¯ÃÂ© parallel Ã¢ÂÂ modifies current division, not correct.
Option B: 125Ã¢ÂÂ¯ÃÂ© series Ã¢ÂÂ too large.
Option C: 25Ã¢ÂÂ¯ÃÂ© parallel Ã¢ÂÂ wrong effect.
Option D: 25Ã¢ÂÂ¯ÃÂ© series Ã¢ÂÂ Ã¢ÂÂ correct.
Final Answer:
$ \boxed{25 \ \Omega 	ext{ resistance in series (Option D)}} $

Answer

$125 \Omega$ resistance in parallel

Answer

$125 \Omega$ resistance in series

Answer

$25 \Omega$ resistance in parallel

Answer

$25 \Omega$ resistance in series

Current Electricity

The voltmeter has range 10 V and its internal resistance is 50 \Omega. To increase the range of voltmeter to 15 V , the resistance which is to be connected is

Choose the correct answer: