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`
Advertisements
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`
RELATED QUESTIONS
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.)
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.
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.
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 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.
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.
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?
If n cells each of emf e and internal resistance r are connected in parallel, then the total emf and internal resistance will be ______.
A straight line plot showing the terminal potential difference (V) of a cell as a function of current (I) drawn from it, is shown in the figure. The internal resistance of the cell would be then ______.

