मराठी

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2015-2016 (March) All India Set 2 C

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

A cell of emf 'E' and internal resistance 'r' is connected across a variable resistor 'R'. Plot a graph showing variation of terminal voltage 'V' of the cell versus the current 'I'. Using the plot, show how the emf of the cell and its internal resistance can be determined.


The earth’s surface has a negative surface charge density of 10−9 C m−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)


Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?


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

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

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


Consider N = n1n2 identical cells, each of emf ε and internal resistance r. Suppose n1 cells are joined in series to form a line and n2 such lines are connected in parallel.

The combination drives a current in an external resistance R. (a) Find the current in the external resistance. (b) Assuming that n1 and n2 can be continuously varied, find the relation between n1, n2, R and r for which the current in R is maximum.


A coil of resistance 100 Ω is connected across a battery of emf 6.0 V. Assume that the heat developed in the coil is used to raise its temperature. If the heat capacity of the coil is 4.0 J K−1, how long will it take to raise the temperature of the coil by 15°C?


Find the emf of the battery shown in the figure:


Answer the following question.
What is the end error in a meter bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = Ω and S respectively.  When the resistance S is shunted with equal resistance, the new balance length found to be 1.5 l1, where l2 is the initial balancing length. calculate the value of s.


If n cells each of emf e and internal resistance r are connected in parallel, then the total emf and internal resistance will be ______.


Five cells each of emf E and internal resistance r send the same amount of current through an external resistance R whether the cells are connected in parallel or in series. Then the ratio `("R"/"r")` is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×