Physics

Physics

Problem

A 9 V9\,\text{V} battery is connected to a 9 Ω9\,\Omega resistor. By mistake, a voltmeter of resistance 9 kΩ9\,\text{k}\Omega is inserted in series with the resistor instead of being connected across it. Approximately what reading will the voltmeter show? a) Nearly 4.5 V4.5\,\text{V} b) Nearly 8.1 V8.1\,\text{V} c) Nearly 9 V9\,\text{V} d) Nearly 0 V0\,\text{V}
Intuitive explanation

The key insight is that a voltmeter measures the voltage across itself, not across the resistor—and when inserted in series, it becomes part of the circuit. You need to find what fraction of the battery's voltage drops across the voltmeter using the voltage divider rule. Since the voltmeter's resistance is much larger than the load resistor, it will "steal" most of the voltage, giving you a reading close to the full battery voltage.

Intuitive explanation

The key insight is that a voltmeter measures the voltage across itself, not across the resistor—and when inserted in series, it becomes part of the circuit. You need to find what fraction of the battery's voltage drops across the voltmeter using the voltage divider rule. Since the voltmeter's resistance is much larger than the load resistor, it will "steal" most of the voltage, giving you a reading close to the full battery voltage.

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