Advertisements
Advertisements
प्रश्न
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 ______.
विकल्प
`((E - V)/E)R`
`((E - V)/R)`
`((E - V)R)/I`
`((E - V)/V)R`
Advertisements
उत्तर
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`
संबंधित प्रश्न
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.
Six lead-acid types of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
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.
Plot a graph showing variation of voltage vs the current drawn from the cell. How can one get information from this plot about the emf of the cell and its internal resistance?
Find the equivalent resistance of the network shown in the figure between the points a and b.
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.
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:
The internal resistance of a cell is the resistance of ______
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)
