English

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. - Physics

Advertisements
Advertisements

Question

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 ______.

Options

  • `((E - V)/E)R`

  • `((E - V)/R)`

  • `((E - V)R)/I`

  • `((E - V)/V)R`

MCQ
Fill in the Blanks

Solution

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 `underlinebb(((E - V)/V)R)`.

Explanation:

The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it. It is measured in volts (V). It is equal to the potential difference across the terminals of the cell when no current is flowing through it. E = I(R + r), where E = electromotive force in volts, I = current in amperes, R = resistance of the load in the circuit in ohms, and r = internal resistance of the cell in ohms. That is, E = IR + Ir, E = V + Ir.

Rearranging the equation we get,

r = `((E - V))/I`

Substituting `I = V/R` in the above equation, we get,

r = `((E - V)R)/V`

∴ Internal resistance, r = `((E - V)R)/V`

shaalaa.com
  Is there an error in this question or solution?
2022-2023 (March) Outside Delhi Set 1

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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?


Two identical cells of emf 1.5 V each joined in parallel, supply energy to an external circuit consisting of two resistances of 7 Ω each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell.


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.


A potentiometer wire of length 1.0 m has a resistance of 15 Ω. It is connected to a 5 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.


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 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.


The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0°C and 0.001°C. Find the thermo-emf (Seebeck emf) developed. For bismuth-silver, a = − 46 × 10−6 V°C−1 and b = −0.48 × 10−6 V°C−2.


An energy source will supply a constant current into the load if its internal resistance is ______.

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:


Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R.

Obtain expressions for (i) current flowing in the circuit, and (ii) the terminal potential differences across the equivalent cell.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×