A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 ohm resi

Q: Why do we connect a high resistance in series to make a voltmeter?

A voltmeter is connected in parallel to the component whose voltage is being measured. If its resistance were low, it would draw a significant amount of current, altering the circuit's behavior. A high series resistance ensures it draws negligible current.

Q: Why is a shunt connected in parallel to make an ammeter?

An ammeter is connected in series to measure the total current. A galvanometer can only handle a tiny current ($i_g$). The parallel shunt acts as a bypass, allowing the majority of the large current to flow through it while keeping the galvanometer safe.

Q: What exactly is $i_g$?

It is the full-scale deflection current. This is the maximum safe current that can flow through the galvanometer coil to make the needle point to the maximum value on its scale.

Q: Why is the voltage across the galvanometer and the shunt the same?

Because they are connected in parallel! In any parallel circuit configuration, the potential difference across all parallel branches is always identical.

Q: How did we get $(1.5 - 0.006)$ for the shunt current?

According to Kirchhoff's Current Law, the total current entering the junction is $1.5\text{ A}$. Since $0.006\text{ A}$ goes through the galvanometer, the remaining current, which is $1.5 - 0.006$, must flow through the shunt.

Question

A galvanometer gives full scale deflection with $0.006	ext{ A}$ current. By connecting it to a $4990\ \Omega$ resistance, it can be converted into a voltmeter of range $0	ext{-}30	ext{ V}$. If connected to a $\frac{2n}{249}\ \Omega$ resistance, it becomes an ammeter of range $0	ext{-}1.5	ext{ A}$. The value of $n$ is

Sigma Percentile · Accepted Answer

Have you ever wondered how a single, delicate instrument like a galvanometer can be transformed to measure both high voltages and large currents? It all comes down to the strategic placement of resistors. Let's dive into this fascinating problem and uncover the mechanics behind these conversions.

### The Voltmeter Conversion: Building a High-Resistance Path

A galvanometer is highly sensitive; it gives a full-scale deflection with just a tiny current—in our case, $0.006	ext{ A}$. If we want it to measure a large voltage, say up to $30	ext{ V}$, we cannot connect it directly across the voltage source. It would draw too much current and burn out!

To prevent this, we connect a very high resistance in series with the galvanometer. This ensures that even at the maximum voltage, only the safe, full-scale deflection current $i_g$ flows through the circuit. The governing equation for this series combination is:

$$V = i_g(G + R)$$

Here, $V$ is the maximum voltage, $i_g$ is the full-scale current, $G$ is the galvanometer's internal resistance, and $R$ is the series resistance. Substituting our known values:

$$30 = 0.006(G + 4990)$$

By dividing $30$ by $0.006$, we get $5000$. This means the total resistance of the circuit must be $5000\ \Omega$.

$$G + 4990 = 5000$$

Subtracting the series resistance, we elegantly find the internal resistance of the galvanometer:

$$G = 10\ \Omega$$

### The Ammeter Conversion: Creating a Bypass Lane

Now, what if we want to measure a large current, up to $1.5	ext{ A}$? Again, feeding this directly into our delicate galvanometer would be disastrous. Instead, we provide a 'bypass lane' for the excess current by connecting a very small resistance, called a **shunt** ($S$), in parallel with the galvanometer.

In a parallel circuit, the potential difference across both branches is identical. Therefore, the voltage across the galvanometer equals the voltage across the shunt:

$$i_g G = (I - i_g) S$$

Here, $I$ is the total current ($1.5	ext{ A}$), and $(I - i_g)$ is the current flowing through the shunt. Let's plug in our values:

$$0.006 	imes 10 = (1.5 - 0.006) S$$

This simplifies to:

$$0.06 = 1.494 S$$

Solving for the shunt resistance $S$:

$$S = \frac{0.06}{1.494} = \frac{60}{1494}\ \Omega$$

### The Final Calculation

The problem states that this shunt resistance is equal to $\frac{2n}{249}\ \Omega$. We simply equate our calculated value to this expression:

$$\frac{2n}{249} = \frac{60}{1494}$$

To isolate $n$, we cross-multiply:

$$2n = \frac{60 	imes 249}{1494}$$

Notice how beautifully the numbers align! $249$ goes into $1494$ exactly $6$ times. So, we have:

$$2n = \frac{60}{6} = 10$$

Dividing by $2$, we arrive at our final answer:

$$n = 5$$

Through this journey, we've seen how Ohm's law and parallel/series circuit principles allow us to manipulate a simple galvanometer into versatile measuring instruments. It's a perfect example of how theoretical physics translates into practical engineering!

Enter Numerical Value:

Visualized Solution

$V o l t m e t er C o n v er s i o n$

$V = i_{g} (G + R)$

$S u b s t i t u t i o n$

$G + 4990 = 5000$

$G = 10 Ω$

$A mm e t er C o n v er s i o n$

$i_{g} G = (I - i_{g}) S$

$S u b s t i t u t i o n$

$0.06 = 1.494 S$

$S = \frac{60}{1494}$

$n = 5$

$T h e W a y F or w a r d$

The Sigma Insight: Electrical Instruments

The Voltmeter Conversion

The Ammeter Conversion

The Final Calculation

Similar Questions

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A galvanometer having a coil resistance $100Ω$ gives a full scale deflection when a current of $1 mA$ is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of $10 V$ ?

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A microammeter has a resistance of $100 Ω$ and full scale range of $50 μ A$ . It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s)

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=ig​(G+R)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G+4990=5000

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G=10 Ω

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ig​G=(I−ig​)S

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0.06=1.494S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=149460​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=5

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

The Sigma Insight: Electrical Instruments

The Voltmeter Conversion

The Ammeter Conversion

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

$V o l t m e t er C o n v er s i o n$

$V = i_{g} (G + R)$

$S u b s t i t u t i o n$

$G + 4990 = 5000$

$G = 10 Ω$

$A mm e t er C o n v er s i o n$

$i_{g} G = (I - i_{g}) S$

$S u b s t i t u t i o n$

$0.06 = 1.494 S$

$S = \frac{60}{1494}$

$n = 5$

$T h e W a y F or w a r d$

In the circuit, the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance, the value of the resistor $R$ will be