हिंदी

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

प्रश्न

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

उत्तर

(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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2015-2016 (March) All India Set 2 C

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?


A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.


A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph showing the variation of terminal potential ‘V’ with resistance R. Predict from the graph the condition under which ‘V’ becomes equal to ‘E’.


Two non-ideal batteries are connected in series. Consider the following statements:-

(A) The equivalent emf is larger than either of the two emfs.

(B) The equivalent internal resistance is smaller than either of the two internal resistances.


How many time constants will elapse before the power delivered by a battery drops to half of its maximum value in an RC circuit?


Find the emf of the battery shown in the figure:


A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, versus R.


A cell E1 of emf 6 V and internal resistance 2 Ω is connected with another cell E2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is ______.


A battery of EMF 10V sets up a current of 1A when connected across a resistor of 8Ω. If the resistor is shunted by another 8Ω resistor, what would be the current in the circuit? (in A)


An ac generator generates an emf which is given by e = 311 sin (240 πt) V. Calculate:

  1. frequency of the emf.
  2. r.m.s. value of the emf.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×