Two resistors, 400 ohm and 800 ohm are connected in series with a 6 V battery. It is desired to meas

Q: Why doesn't the ammeter read exactly $6 / 1200 = 5\ \text{mA}$?

Because a real ammeter has its own internal resistance. When placed in series, it adds to the total resistance of the circuit, slightly decreasing the current it is trying to measure.

Q: Why is the voltmeter connected in parallel?

A voltmeter measures potential difference between two points. To measure the voltage across a resistor, it must be connected across those same two points, which means it is in parallel with the resistor.

Q: Does the voltmeter change the voltage it's measuring?

Yes! By providing an alternate path for the current, it lowers the equivalent resistance of that section, which in turn changes the total current and the voltage drops across all components.

Q: What would happen if the voltmeter was ideal?

An ideal voltmeter has infinite resistance. It would draw zero current, the parallel resistance would remain $400\ \Omega$, and it would measure exactly $2\ \text{V}$.

Q: How do we minimize these measurement errors?

To minimize errors, an ammeter should have a resistance as close to zero as possible, and a voltmeter should have a resistance as close to infinity as possible compared to the circuit components.

Question

Two resistors, $400 \Omega$ and $800 \Omega$ are connected in series with a $6 	ext{ V}$ battery. It is desired to measure the current in the circuit. An ammeter of $10 \Omega$ resistance is used for this purpose. What will be the reading in the ammeter? Similarly, if a voltmeter of $1000 \Omega$ resistance is used to measure the potential difference across the $400 \Omega$ resistor, what will be the reading in the voltmeter?

Sigma Percentile · Accepted Answer

### The Ideal vs. The Real World

When we first learn about circuits, we often assume our measuring instruments are perfect. We imagine ammeters with zero resistance and voltmeters with infinite resistance.

But in the real world, instruments are physical devices that interact with the circuit they are measuring. This problem is a beautiful demonstration of the **Observer Effect** in electronics: the act of measuring a circuit inevitably changes it.

Let's dive into how these real-world imperfections alter our expected results.

### Measuring Current: The Ammeter's Toll

We start with a simple series circuit: a $6\ 	ext{V}$ battery driving current through a $400\ \Omega$ and an $800\ \Omega$ resistor.

To measure this current, we insert an ammeter in series. However, this ammeter has its own internal resistance of $10\ \Omega$. Because it is in series, this resistance adds directly to the total resistance of the circuit.

$$R_{	ext{eq}} = R_1 + R_2 + R_A$$

$$R_{	ext{eq}} = 400 + 800 + 10 = 1210\ \Omega$$

If the ammeter were ideal, the total resistance would just be $1200\ \Omega$. The extra $10\ \Omega$ acts like a tiny toll booth, slightly restricting the flow of electrons.

Now, we use Ohm's Law to find the actual current flowing through this modified circuit:

$$i = \frac{V}{R_{	ext{eq}}} = \frac{6}{1210} \approx 4.96 	imes 10^{-3}\ 	ext{A}$$

So, the ammeter will read **$4.96\ 	ext{mA}$**. Notice how this is slightly less than the ideal $5\ 	ext{mA}$ we would expect without the ammeter's interference!

### Measuring Voltage: The Voltmeter's Burden

Next, we remove the ammeter and attempt to measure the potential difference across the $400\ \Omega$ resistor. To do this, we connect a voltmeter in parallel with it.

This voltmeter has a resistance of $1000\ \Omega$. By placing it in parallel, we are essentially giving the current a second path to flow through. This reduces the overall resistance of that section of the circuit.

Let's calculate the equivalent resistance of this parallel combination:

$$R_p = \frac{R_1 R_V}{R_1 + R_V}$$

$$R_p = \frac{400 	imes 1000}{400 + 1000} = \frac{400000}{1400} \approx 285.71\ \Omega$$

The resistance of that section has dropped from $400\ \Omega$ to $285.71\ \Omega$! This changes the entire dynamics of the circuit. The new total resistance is:

$$R_{	ext{eq}}' = R_p + R_2 = 285.71 + 800 = 1085.71\ \Omega$$

Because the total resistance has decreased, the main current drawn from the battery will increase:

$$i' = \frac{V}{R_{	ext{eq}}'} = \frac{6}{1085.71} \approx 5.53 	imes 10^{-3}\ 	ext{A}$$

Finally, the voltmeter reads the voltage across its own parallel section. We find this by multiplying the new main current by the parallel equivalent resistance:

$$V_{	ext{read}} = i' 	imes R_p = (5.53 	imes 10^{-3}) 	imes 285.71 \approx 1.58\ 	ext{V}$$

### The Grand Takeaway

The voltmeter reads **$1.58\ 	ext{V}$**. If we had used an ideal voltmeter, the reading would have been exactly $2\ 	ext{V}$.

This problem perfectly illustrates why high-quality voltmeters are designed to have extremely high resistances (often in the megaohms), and ammeters are designed to have extremely low resistances. The closer they are to ideal, the less they disturb the delicate electrical ecosystem they are trying to measure!

Visualized Solution

\text{Circuit Analysis}

\text{Measuring Current}

R_{\text{eq}} = R_1 + R_2 + R_A

i = \frac{V}{R_{\text{eq}}}

\text{Measuring Voltage}

R_p = \frac{R_1 R_V}{R_1 + R_V}

R_{\text{eq}}' = R_p + R_2

i' = \frac{V}{R_{\text{eq}}'}

V_{\text{read}} = i' \times R_p

\text{Final Answers}

The Sigma Insight: Electrical Instruments

The Ideal vs

Measuring Current

Measuring Voltage

The Grand Takeaway

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