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Current Electricity - Cells, EMF and Internal Resistance

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electromotive Force (EMF) ε\varepsilon: It is the maximum potential difference between the two electrodes of a cell when no current is drawn from the cell (open circuit).

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Internal Resistance rr: The resistance offered by the electrolyte and electrodes inside the cell to the flow of current. It depends on the nature of the electrolyte, distance between electrodes, and the area of electrodes immersed.

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Terminal Potential Difference VV: The potential difference across the terminals of a cell when current II is flowing through the external circuit. For a discharging cell, V<εV < \varepsilon and V=ε−IrV = \varepsilon - Ir.

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Charging a Cell: When a cell is being charged by an external source, the current flows in the opposite direction, and the terminal potential difference becomes V=ε+IrV = \varepsilon + Ir.

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Series Combination: For nn cells connected in series, the equivalent EMF is εeq=ε1+ε2+...+εn\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2 + ... + \varepsilon_n and the equivalent internal resistance is req=r1+r2+...+rnr_{eq} = r_1 + r_2 + ... + r_n.

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Parallel Combination: For cells connected in parallel, the equivalent internal resistance reqr_{eq} is given by 1req=∑1ri\frac{1}{r_{eq}} = \sum \frac{1}{r_i} and the equivalent EMF εeq\varepsilon_{eq} satisfies εeqreq=∑εiri\frac{\varepsilon_{eq}}{r_{eq}} = \sum \frac{\varepsilon_i}{r_i}.

📐Formulae

I=εR+rI = \frac{\varepsilon}{R + r}

V=ε−IrV = \varepsilon - Ir

r=R(εV−1)r = R \left( \frac{\varepsilon}{V} - 1 \right)

εeq=∑i=1nεi (Series connection)\varepsilon_{eq} = \sum_{i=1}^{n} \varepsilon_i \text{ (Series connection)}

req=∑i=1nri (Series connection)r_{eq} = \sum_{i=1}^{n} r_i \text{ (Series connection)}

εeq=ε1r1+ε2r21r1+1r2=ε1r2+ε2r1r1+r2 (Parallel connection for two cells)\varepsilon_{eq} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \text{ (Parallel connection for two cells)}

1req=1r1+1r2 (Parallel connection)\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \text{ (Parallel connection)}

💡Examples

Problem 1:

A storage battery of a car has an EMF of 12 V12\text{ V}. If the internal resistance of the battery is 0.4 Ω0.4\text{ }\Omega, what is the maximum current that can be drawn from the battery?

Solution:

The current II in a circuit is given by I=εR+rI = \frac{\varepsilon}{R + r}. For the current to be maximum, the external resistance RR must be zero (short circuit). Imax=εr=120.4=30 AI_{max} = \frac{\varepsilon}{r} = \frac{12}{0.4} = 30\text{ A}

Explanation:

Maximum current occurs when there is no external load resistance. In this case, the entire EMF is dropped across the internal resistance of the cell.

Problem 2:

A cell of EMF ε\varepsilon and internal resistance rr is connected across a variable resistor RR. When R=5 ΩR = 5\text{ }\Omega, the current is 0.5 A0.5\text{ A}. When R=11 ΩR = 11\text{ }\Omega, the current is 0.25 A0.25\text{ A}. Find the EMF and internal resistance.

Solution:

Using the formula ε=I(R+r)\varepsilon = I(R + r):

  1. For R=5 ΩR = 5\text{ }\Omega: ε=0.5(5+r)\varepsilon = 0.5(5 + r)
  2. For R=11 ΩR = 11\text{ }\Omega: ε=0.25(11+r)\varepsilon = 0.25(11 + r) Equating both: 0.5(5+r)=0.25(11+r)0.5(5 + r) = 0.25(11 + r) 2(5+r)=11+r2(5 + r) = 11 + r 10+2r=11+r10 + 2r = 11 + r r=1 Ωr = 1\text{ }\Omega Substituting rr in (1): ε=0.5(5+1)=3 V\varepsilon = 0.5(5 + 1) = 3\text{ V}

Explanation:

We use the circuit equation twice with the given conditions to form a system of linear equations for the two unknowns ε\varepsilon and rr.