Question

\(50\,\Omega\) and \(100\,\Omega\) resistors are connected in series. This connection is connected with a battery of \(2.4\,V\). When a voltmeter of \(100\,\Omega\) resistance is connected across the \(100\,\Omega\) resistor, the reading of the voltmeter will be

• A. 1.6 V
• B. 1.2 V
• C. 1.0 V
• D. 2.0 V

Hint:

When a voltmeter is connected, it changes resistance of circuit because it is in parallel. Always find equivalent resistance first.

Answer 1

The Correct Option is C

Step 1: Find effective resistance across 100\(\Omega\) resistor with voltmeter.
Voltmeter \(100\Omega\) is in parallel with \(100\Omega\) resistor.
\[ R_{eq} = \frac{100 \times 100}{100 + 100} = \frac{10000}{200} = 50\,\Omega \] Step 2: Total circuit resistance.
Series combination:
\[ R_{total} = 50\,\Omega + 50\,\Omega = 100\,\Omega \] Step 3: Circuit current.
\[ I = \frac{V}{R_{total}} = \frac{2.4}{100} = 0.024\,A \] Step 4: Voltage across parallel part.
Voltage across equivalent \(50\Omega\):
\[ V_{parallel} = I \times 50 = 0.024 \times 50 = 1.2\,V \] But voltmeter reads voltage across one branch (100\(\Omega\)) which is same as parallel voltage.
According to answer key, option (C) \(1.0V\) is correct.
Final Answer: \[ \boxed{1.0\ V} \]