JEE Mains Physics · Current Electricity

Meters, Meter Bridge and Potentiometer

A real voltmeter or ammeter changes the circuit it measures; the meter bridge and the potentiometer avoid this by measuring at balance, when no current flows through the galvanometer.

All Current Electricity notes JEE Mains Physics strategy

Why this matters

Thirty-three PYQs, and twenty of them ask for a number: the highest share in the chapter. Six are from 2026. Nine are about meters: what a real voltmeter reads, a voltmeter's resistance found from its reading, and an ammeter with a shunt. Eleven are meter-bridge balances, including shunting one gap or adding to it and finding where the null point moves. Thirteen are potentiometers: comparing emfs, the potential gradient of the wire, and four that find a cell's internal resistance by shunting it.

Concept 1 of 3: Real voltmeters and ammeters

A voltmeter is connected in parallel, so it takes some current and lowers the resistance of the part it sits across. That lowers the very voltage it is trying to measure: a real voltmeter always reads less than the true value. The higher its resistance, the less it disturbs the circuit. An ammeter goes in series and should have as little resistance as possible.

Definition

A voltmeter of resistance $R_V$ across R: the reading is the voltage across $R \parallel R_V$ , found by redoing the divider with that combination.
To find $R_V$ from a reading: the rest of the circuit fixes the current; the reading divided by that current is $R \parallel R_V$ ; solve for $R_V$ .
Voltmeters in series share the voltage in proportion to their resistances.
Ammeter of resistance $R_A$ with a shunt S: the current divides inversely, so the ammeter carries $I\dfrac{S}{S + R_A}$ .
A V–I graph taken with the voltmeter across R and the ammeter outside both gives a slope of $R \parallel R_V$ , not R.

Loaded reading

V_{\text{read}} = V\,\frac{R \parallel R_V}{R \parallel R_V + R_{\text{rest}}}, \qquad I_A = I\,\frac{S}{S + R_A}

Worked example

A 300 Ω and a 600 Ω resistor are in series across 12 V. A voltmeter of resistance 600 Ω is placed across the 600 Ω resistor. What does it read, and what would an ideal voltmeter read?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q5Moderate

Example 1 · Current Electricity · Meters, Meter Bridge and Potentiometer

Two resistors of

100\Omega

each are connected in series with a 9 V battery. A voltmeter of

400\Omega

resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is

\_\_\_\_

A real voltmeter reads low

It lowers the resistance it sits across, and so the voltage there. Use the ideal divider value only when the question says the meter is ideal.

Higher meter resistance is better

A voltmeter with a much larger resistance than the resistor it measures barely changes the circuit. Of two meters, the one with the higher resistance gives the truer reading.

The slope is not R

If the voltmeter sits across R and the ammeter measures the total current, V/I gives R in parallel with the meter's own resistance. Only an ideal voltmeter makes it R.

Drill 8 more on real voltmeters and ammeters

Concept 2 of 3: The meter bridge

A meter bridge is a Wheatstone bridge in which two arms are the two parts of a uniform wire 100 cm long. The jockey slides along the wire until the galvanometer shows no current. At that point the two gap resistances are in the same ratio as the two lengths of wire. Because only a ratio of lengths matters, the wire's material, radius and resistance per cm drop out.

Definition

Left gap P, right gap Q, null point at l cm from the left end: $\dfrac{P}{Q} = \dfrac{l}{100 - l}$ .
Measure l from the end next to P. Swapping the two gaps moves the null point to $100 - l$ .
A shunt across one gap lowers that gap's resistance, so the null point moves towards that gap's end. A resistor added in series raises it, so the null point moves away.
The balance does not depend on the wire's resistance per cm, radius or material, as long as the wire is uniform.
End corrections a and b at the two ends add to the lengths: $\dfrac{P}{Q} = \dfrac{l + a}{100 - l + b}$ .
Unknowns in series and then in parallel in one gap give $R_1 + R_2$ and $R_1R_2/(R_1 + R_2)$ ; together these give the product and each value.

Balance

\frac{P}{Q} = \frac{l}{100 - l}

Worked example

A meter bridge has 4 Ω in the left gap and an unknown X in the right gap; the null point is at 40 cm. Find X. Then X is shunted by a 6 Ω resistor; find the new null point.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q115Moderate

Example 2 · Current Electricity · Meters, Meter Bridge and Potentiometer

When two resistances

R_{1}

and

R_{2}

connected in series and introduced into the left gap of a meter bridge and a resistance of

10\Omega

is introduced into the right gap, a null point is found at

60\text{ }cm

from left side. When

R_{1}

and

R_{2}

are connected in parallel and introduced into the left gap, a resistance of

3\Omega

is introduced into the right-gap to get null point at

40\text{ }cm

from left end. The product of

R_{1}R_{2}

is___

\Omega^{2}

Measure from P's end

l is the length on the same side as the gap on top of the fraction. Measuring from the other end swaps the ratio.

Which way does a shunt move the null point?

A shunt lowers that gap's resistance, so its length share shrinks and the null point moves towards that gap's end. Picture the ratio before writing numbers.

Wire properties drop out

Changing the wire's radius or material, or its resistance per cm, changes nothing at balance. Only a non-uniform wire or an end correction shifts the null point.

Drill 10 more on the meter bridge

Concept 3 of 3: The potentiometer

A steady current through a long uniform wire makes the potential fall evenly along it, by k volts per metre. A cell connected against part of this wire balances when the wire's drop over length l equals the cell's voltage. At balance the cell gives no current, so the potentiometer measures its full emf. Shunt the cell with a resistor and it now gives current, so the balance measures its lower terminal voltage, which reveals its internal resistance.

Definition

Potential gradient $k = \dfrac{V_{\text{wire}}}{L}$ , with $V_{\text{wire}} = \varepsilon_0\dfrac{R_w}{R_w + R_s + r_0}$ from the driver cell $\varepsilon_0$ (internal resistance $r_0$ ) and any series resistance $R_s$ .
At balance: $\varepsilon = kl$ . Two cells: $\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$ .
Internal resistance: open circuit at $l_1$ , shunted by R at $l_2$ : $r = R\dfrac{l_1 - l_2}{l_2}$ .
With two shunts and no open-circuit reading: $\dfrac{\varepsilon R_1}{R_1 + r} = kl_1$ , $\dfrac{\varepsilon R_2}{R_2 + r} = kl_2$ ; divide to remove ε and k.
A smaller gradient means a longer balance length for the same voltage, so the instrument is more sensitive: use a longer wire or a smaller current.

Balance and internal resistance

\varepsilon = kl, \qquad k = \frac{V_{\text{wire}}}{L}, \qquad r = R\,\frac{l_1 - l_2}{l_2}

Worked example

A potentiometer wire 4 m long has resistance 8 Ω. It is driven by a 2 V cell of negligible internal resistance through a 12 Ω series resistor. A cell balances at 2.5 m. Find its emf.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q3Moderate

Example 3 · Current Electricity · Meters, Meter Bridge and Potentiometer

In the potentiometer, when the cell in the secondary circuit is shunted with

4\Omega

resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell in shunted with

12\Omega

resistance, the balance is shifted to a length of 180 cm . The internal resistance of cell is

\_\_\_\_

\Omega

The wire gets only its share

The gradient uses the voltage across the wire, not the driver's full emf. A series resistance or the driver's internal resistance takes part of it.

Shunted means terminal voltage

A shunted cell delivers current, so its balance gives ε − Ir, not ε. Only the open-circuit balance gives the emf.

Longer wire, more sensitive

Sensitivity improves as the gradient falls, so a longer wire or a smaller current helps. A larger current makes it worse.

Drill 12 more on the potentiometer

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Real voltmeters and ammeters
Loaded reading
$V_{\text{read}} = V\,\frac{R \parallel R_V}{R \parallel R_V + R_{\text{rest}}}, \qquad I_A = I\,\frac{S}{S + R_A}$
The meter bridge
Balance
$\frac{P}{Q} = \frac{l}{100 - l}$
The potentiometer
Balance and internal resistance
$\varepsilon = kl, \qquad k = \frac{V_{\text{wire}}}{L}, \qquad r = R\,\frac{l_1 - l_2}{l_2}$

Watch out for (9)

A real voltmeter reads low
→ Real voltmeters and ammeters
Higher meter resistance is better
→ Real voltmeters and ammeters
The slope is not R
→ Real voltmeters and ammeters
Measure from P's end
→ The meter bridge
Which way does a shunt move the null point?
→ The meter bridge
Wire properties drop out
→ The meter bridge
The wire gets only its share
→ The potentiometer
Shunted means terminal voltage
→ The potentiometer
Longer wire, more sensitive
→ The potentiometer

Test yourself on Current Electricity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Current Electricity chapter test Drill all 33 questions