English

A Battery of Emf 12 V and Internal Resistance 2 Ω is Connected to a 4 Ω 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. - Physics

Advertisements
Advertisements

Question

A battery of emf 12 V and internal resistance 2 Ω is connected to a 4 Ω 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?

Advertisements

Solution

(i) According to the definition of terminal potential difference,

E = V + Ir

`=>V=E-Ir`

E = 12 V, r = 2 Ω

V = 12 – 2I

When the voltmeter is connected across the cell.

`I=12/(4+2)=2A`

V1 = 12 – 2 (2) = 8 V

When the voltmeter is connected across the resistor.

V2 = IR

= 2 × 4 = 8 V

V1 = V2

Hence, from the above relation we can see that when voltmeter placed across the cell and across the resistor, gives the same reading

(ii) Voltmeter has very high resistance to ensure that it's connection does not alter the flow of current in the circuit. We connect it in parallel and we also know that current chooses only the low resistance path. Hence, it is connected in parallel to the load across which potential difference is to be measured.

Ammeter measures value of current flowing through the circuit so it should be connected in the series. Ammeter has very low resistance to ensure that all the current flows through it. Thus, it gives a correct reading of the current when connected in series.

shaalaa.com
  Is there an error in this question or solution?
2015-2016 (March) All India Set 2 C

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Two cells of emfs 1.5 V and 2.0 V,  having internal resistances 0.2 Ω and 0.3 Ω, respectively, are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.


In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?


Nichrome and copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more? Justify your answer.


In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.


Two cells of emf E1, E2 and internal resistance r1 and r2 respectively are connected in parallel as shown in the figure.

Deduce the expressions for

(1) the equivalent e.m.f of the combination

(2) the equivalent resistance of the combination, and

(3) the potential difference between the point A and B.


Can the potential difference across a battery be greater than its emf?


Apply the first law of thermodynamics to a resistor carrying a current i. Identify which of the quantities ∆Q, ∆U and ∆W are zero, positive and negative.


Do all thermocouples have a neutral temperature?


The internal resistance of a cell is the resistance of ______


A cell of emf E is connected across an external resistance R. When current 'I' is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance 'r' of the cell is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×