The actual value of resistance R, shown in the figure is 30 ohms. This is measured in an experiment

Q: Why is the measured resistance less than the actual resistance?

Because the voltmeter is connected in parallel, it provides an additional path for the current. This reduces the overall equivalent resistance of that section of the circuit.

Q: What is an ideal voltmeter?

An ideal voltmeter has infinite internal resistance. It draws zero current, meaning it measures the potential difference without altering the circuit's behavior.

Q: Why do we use the parallel resistance formula here?

The voltmeter is connected across the resistor to measure its voltage drop. Components connected across the same two points are in parallel, so their equivalent resistance follows the parallel formula.

Q: How did we get 28.5 ohms?

The problem states the measured value is 5% less than the actual 30 ohms. 5% of 30 is 1.5, and subtracting that from 30 gives 28.5 ohms.

Q: What is the 'loading effect'?

It's the alteration of a circuit's voltage or current caused by connecting a measuring instrument. Here, the voltmeter 'loads' the circuit by drawing current, causing a measurement error.

Question

The actual value of resistance $R$, shown in the figure is $30\,\Omega$. This is measured in an experiment as shown using the standard formula $R = \frac{V}{I}$, where $V$ and $I$ are the readings of the voltmeter and ammeter, respectively. If the measured value of $R$ is $5\%$ less, then the internal resistance of the voltmeter is

Options:

(A) $600\,\Omega$

(B) $570\,\Omega$

(C) $350\,\Omega$

(D) $35\,\Omega$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are setting up an experiment to measure the resistance of a given resistor, $R$. You connect an ammeter in series to measure the current and a voltmeter in parallel to measure the voltage drop.

According to Ohm's law, the resistance should simply be $R = \frac{V}{I}$. However, there is a catch! An ideal voltmeter has infinite internal resistance, meaning it draws absolutely zero current. But in the real world, every voltmeter has some finite internal resistance, let's call it $R_V$.

Because the voltmeter is connected in parallel with the resistor $R$, it provides an alternative path for the current. The ammeter doesn't just measure the current flowing through $R$; it measures the total current flowing through both $R$ and the voltmeter.

### The Master Equation

This means the resistance you calculate using $R' = \frac{V}{I}$ is not the actual resistance $R$, but rather the **equivalent resistance** of the parallel combination of $R$ and $R_V$.

We know from circuit theory that the equivalent resistance of two resistors in parallel is given by:
$R' = \frac{R \cdot R_V}{R + R_V}$

The problem states that the actual resistance is $R = 30\,\Omega$, but our measured value $R'$ is $5\%$ less than the actual value due to this loading effect.

Let's calculate exactly what this measured value is:
$R' = 30 - \left(\frac{5}{100} 	imes 30ight)$

$R' = 30 - 1.5 = 28.5\,\Omega$

### Final Calculation

Now we have a clear mathematical bridge. We know the theoretical expression for the measured resistance, and we know its numerical value. Let's equate them to find the hidden internal resistance of the voltmeter, $R_V$.

$\frac{30 \cdot R_V}{30 + R_V} = 28.5$

To solve for $R_V$, we cross-multiply:
$30 R_V = 28.5(30 + R_V)$

Expanding the bracket on the right side:
$30 R_V = 855 + 28.5 R_V$

Now, let's group the $R_V$ terms on one side:
$30 R_V - 28.5 R_V = 855$

$1.5 R_V = 855$

Finally, dividing by $1.5$ gives us our answer:
$R_V = \frac{855}{1.5} = 570\,\Omega$

**The internal resistance of the voltmeter is $570\,\Omega$.** This perfectly illustrates why a good voltmeter must have a very high internal resistance—to minimize the current it draws and keep the measured value as close to the true value as possible!

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Visualized Solution

$Circuit Analysis$

$Equivalent Resistance$

$Measured Value R^{'}$

$Calculating R^{'}$

$Equating the Resistances$

$Solving for R_{V}$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Measured Value R′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating R′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating the Resistances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for RV​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward