Q. 10 4.0( 5 Votes )

A battery of emf 12V and internal resistance 2 Ω is connected to a 4 resistor as shown in the figure.

(a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.

(b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit?

Answer :

(a) Now when voltmeter is connected in parallel with the cell it will measure Voltage or Potential Difference across the cell, suppose the current in the circuit is I and the potential difference across the cell at any instant is V volts

Whose emf is E = 12V ,

moreover, internal resistance is r = 2 Ω

moreover, connected across external resistance R = 4 Ω

The circuit is as shown in the figure

Now we know according to ohm’s law, the potential difference across a resistor is equal to product of current flowing through it and its Resistance Mathematically,

V = IR


I = V/R

Where V is the potential difference across the ends of the resistor, I is the current flowing through it, and R is the resistance

Now since internal resistance of cell and external resistance are both connected in series so net resistance of the system will be a summation of the two resistances

R’ = r + R

So net resistance is

R’ = 4 Ω + 2 Ω = 6 Ω

Now potential difference across the combination of resistances will be equal to the Emf of cell

E = 12 V

Now net current in the circuit will be

I = E/R’

Where I is the current in the circuit, E is emf of the cell, R’ is total resistance

So current is

I = 12 V/6 Ω = 2 A

Now potential difference across the internal resistance of cell is

V1 = Ir

Where I is the current flowing through the circuit and ris the internal resistance

In any resistance current flow from higher potential to lower potential, so polarity of voltages of the cell is as shown

If we let the net potential difference across battery be V so

V = E – V1


V = E – Ir

Putting values of E, I , r we get potential difference across cell as

V = 12 V – 2A × 2 Ω = 12 V – 4 V = 8 V

So potential difference across the cell is 8 V so reading of Voltmeter will be 8 V

Now when Voltmeter is connected in parallel across the 4 Ω resistor

As shown in figure

now again current in the circuit will be

I = 2 A

And resistance of the resistor is

R = 4 Ω

now using ohms law potential difference across resistor V will be

V = IR

V = 2A × 4Ω = 8 V

So clearly we can see voltage is the same (8 V) across both cell and 4Ω resistance

(b) In an Electric circuit magnitude of current flowing across two components connected in series is always same and , for any two components connected in parallel voltage across them is same, so an ammeter is connected in series so that same amount of current will pass through it as is passing through other component connected in series with it, so it will measure the same amount of current that is flowing in the circuit or component, now Voltmeter is connected in parallel so that potential difference across voltmeter should be the same as across the component to be measured ,

a simple circuit has been shown showing the connection of voltmeter and ammeter

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HC Verma - Concepts of Physics Part 2