Question:
An ammeter, voltmeter and a resistor are connected in series to a cell and the readings are noted as $I$ and $V$. If another resistor $R$ is connected in parallel with voltmeter, then
An ammeter, voltmeter and a resistor are connected in series to a cell and the readings are noted as $I$ and $V$. If another resistor $R$ is connected in parallel with voltmeter, then
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Parallel addition always reduces resistance → increases current.
Updated On: May 2, 2026
- $I$ and $V$ increase
- $I$ increases
- $I$ and $V$ remain same
- $I$ decreases
- $I$ remains constant
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The Correct Option is B
Solution and Explanation
Concept:
Voltmeter has very high resistance and is ideally connected in parallel.
Step 1: Initial condition:
• Voltmeter in series → behaves like large resistance
• Current is relatively small
Step 2: Adding resistor in parallel with voltmeter:
• Equivalent resistance of parallel combination decreases \[ R_{\text{eq}} = \frac{R \cdot R_v}{R + R_v} \] Since $R_v$ is large: \[ R_{\text{eq}} < R_v \]
Step 3: Effect on total circuit:
• Total resistance of circuit decreases
Step 4: Using Ohm’s law: \[ I = \frac{V}{R} \] Decrease in resistance → increase in current.
Step 5: Effect on voltmeter reading:
• Current redistribution changes potential drop
• But key effect asked is current increase Final Conclusion: Current increases.
Step 1: Initial condition:
• Voltmeter in series → behaves like large resistance
• Current is relatively small
Step 2: Adding resistor in parallel with voltmeter:
• Equivalent resistance of parallel combination decreases \[ R_{\text{eq}} = \frac{R \cdot R_v}{R + R_v} \] Since $R_v$ is large: \[ R_{\text{eq}} < R_v \]
Step 3: Effect on total circuit:
• Total resistance of circuit decreases
Step 4: Using Ohm’s law: \[ I = \frac{V}{R} \] Decrease in resistance → increase in current.
Step 5: Effect on voltmeter reading:
• Current redistribution changes potential drop
• But key effect asked is current increase Final Conclusion: Current increases.
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