Is the resistance of the resulting voltmeter exactly $10 \text{ k}\Omega$?

Yes, the total resistance of the voltmeter is $G + R = 100 \Omega + 9900 \Omega = 10000 \Omega$ or $10 \text{ k}\Omega$.

JEE Main 2020

LEVELJEE Main

Animated Solution for Physics - Current Electricity: 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 $10 V$ ?

Select Answer:

Visualized Solution

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

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

$R = \frac{V}{I _{g}} - G$

$G + R = \frac{V}{I _{g}}$
$R = \frac{V}{I _{g}} - G$

$Substitution$

$V = 10 V$
$I_{g} = 1 \times 1 0^{- 3} A$
$G = 100Ω$
$R = \frac{10}{1 \times 1 0 ^{- 3}} - 100$

$Calculation$

$\frac{10}{1 0 ^{- 3}} = 10000Ω$
$R = 10000 - 100$

$R = 9900Ω$

$R = 9900Ω$

$R = 9.9 k Ω$

$R = 9.9 k Ω$

$Ammeter Conversion$

$Conversion to Ammeter:$
$S = \frac{I _{g} G}{I - I _{g}}$

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The Sigma Insight: Electrical Instruments

Solution Diagram

Analyzing the Setup A galvanometer is a highly sensitive device designed to detect very small currents

However, to measure larger potential differences, it must be converted into a voltmeter.

If we connect it directly across a high voltage, the large current would instantly burn out its delicate coil. To prevent this and allow it to measure up to

10 V

, we must connect a very high resistance

R

in series with the galvanometer.

The Master Equation

When the series resistance

R

is connected, the total resistance of the voltmeter becomes

(G + R)

, where

G

is the galvanometer's own resistance.

According to Ohm's law, the maximum voltage

V

that this combination can measure corresponds to the maximum current the galvanometer can safely handle, which is its full-scale deflection current

I_{g}

This gives us our master equation:

V = I_{g} (G + R)

We can rearrange this to solve for the unknown series resistance

R

R = \frac{V}{I _{g}} - G

Final Calculation Now, we simply substitute the given values into our rearranged equation

We know the target voltage

V = 10 V

, the full-scale current

I_{g} = 1 mA = 1 0^{- 3} A

, and the galvanometer resistance

G = 100Ω

R = \frac{10}{1 0 ^{- 3}} - 100

R = 10000 - 100

R = 9900Ω

To match the options provided, we convert this to kilo-ohms:

R = 9.9 k Ω

Thus, a

9.9 k Ω

resistance must be connected in series to achieve the desired voltmeter range.

Common Traps & FAQs

Why do we connect the resistance in series and not in parallel?

A voltmeter must have a very high resistance so that it draws negligible current from the circuit it is measuring. Connecting a high resistance in series achieves this.

What happens if we don't connect a series resistance?

The galvanometer has a low resistance and can only handle very small currents (like 1 mA). Connecting it directly across a 10 V source would cause a huge current to flow, burning out the coil.

Why is the full-scale deflection current

I_{g}

used in the calculation?

We want the voltmeter to show its maximum reading (10 V) exactly when the galvanometer shows its maximum reading (full-scale deflection). This ensures the scales match perfectly.

Can we convert this voltmeter back into an ammeter?

Yes, by removing the series resistance and instead connecting a very low resistance (shunt) in parallel with the galvanometer.

Is the resistance of the resulting voltmeter exactly

10 k Ω

Yes, the total resistance of the voltmeter is

G + R = 100Ω + 9900Ω = 10000Ω

10 k Ω

Similar Questions

JEE Main 2017

LEVELJEE Main

When a current of $5 mA$ is passed through a galvanometer having a coil of resistance $15 Ω$ , it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range $0 - 10 V$ is

(A)

2.045 \times 1 0^{3} Ω

(B)

2.535 \times 1 0^{3} Ω

(C)

4.005 \times 1 0^{3} Ω

(D)

1.985 \times 1 0^{3} Ω

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A moving coil galvanometer of resistance $100Ω$ is used as an ammeter using a resistance $0.1Ω$ . The maximum deflection current in the galvanometer is $100 μ A$ . Find the current in the circuit, so that the ammeter shows maximum deflection.

(A)

100.1 mA

(B)

1000.1 mA

(C)

10.01 mA

(D)

1.01 mA

JEE Advanced 2014

LEVELJEE Advanced

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

JEE Main 2020

LEVELJEE Main

A galvanometer of resistance $G$ is converted into a voltmeter of range $0 - 1 V$ by connecting a resistance $R_{1}$ in series with it. The additional resistance that should be connected in series with $R_{1}$ to increase the range of the voltmeter to $0 - 2 V$ will be

(A)

R_{1} + G

(B)

R_{1}

(C)

G

(D)

R_{1} - G

JEE Advanced 2019

LEVELJEE Advanced

Two identical moving coil galvanometer have $10 Ω$ resistance and full scale deflection at $2 μ A$ current. One of them is converted into a voltmeter of $100 mV$ full scale reading and the other into an Ammeter of $1 mA$ full scale current using appropriate resistors. These are then used to measure the voltage and current in the Ohm's law experiment with $R = 1000 Ω$ resistor by using an ideal cell. Which of the following statement(s) is/are correct ?

* Multiple Correct Options

(A)

The measured value of

R

will be

978 Ω < R < 982 Ω

(B)

The resistance of the Voltmeter will be

100 k Ω

(C)

The resistance of the Ammeter will be

0.02 Ω

(round off to

2^{nd}

decimal place)

(D)

If the ideal cell is replaced by a cell having internal resistance of

5 Ω

then the measured value of

R

will be more than

1000 Ω

JEE Advanced 2026

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As shown in the figure, the resistance of a galvanometer $G$ can be found by the half-deflection method. Here the resistance $R_{2}$ is adjusted such that when the key $K$ is closed the deflection in the galvanometer becomes half of the value as compared to when $K$ is open. Half-deflection is obtained at $R_{2} = 4 Ω$ and thus the galvanometer resistance is found to be $6 Ω$ . In this half-deflection condition the current (in mA) through the resistor $R_{1}$ is:

LEVELJEE Main

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

(A)

200 Ω

(B)

100 Ω

(C)

500 Ω

(D)

1000 Ω

JEE Main 2019

LEVELJEE Advanced

A potentiometer wire $A B$ having length $L$ and resistance $12 r$ is joined to a cell $D$ of EMF $ε$ and internal resistance $r$ . A cell $C$ having emf $\frac{ε}{2}$ and internal resistance $3 r$ is connected. The length $A J$ at which the galvanometer as shown in figure shows no deflection is

(A)

\frac{5}{12} L

(B)

\frac{11}{12} L

(C)

\frac{13}{24} L

(D)

\frac{11}{24} L

JEE Main 2021

LEVELJEE Advanced

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 Ω$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

(A)

2.44 μ A

(B)

2.44 mA

(C)

4.87 mA

(D)

4.87 μ A

LEVELJEE Main

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)

* Multiple Correct Options

(A)

50 V

range with

10 k Ω

resistance in series

(B)

10 V

range with

200 k Ω

resistance in series

(C)

5 mA

range with

1 Ω

resistance in parallel

(D)

10 mA

range with

1 Ω

resistance in parallel

Select Answer:

Visualized Solution

.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; } R=Ig​V​−G

.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; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=9900Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=9.9 kΩ

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

The Sigma Insight: Electrical Instruments

Analyzing the Setup A galvanometer is a highly sensitive device designed to detect very small currents

The Master Equation

Final Calculation Now, we simply substitute the given values into our rearranged equation

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

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

$R = \frac{V}{I _{g}} - G$

$Substitution$

$Calculation$

$R = 9900Ω$

$R = 9.9 k Ω$

$Ammeter Conversion$

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