A cell of emf 2V and internal resistance 0-1 Omega- is connected to a 3-9 Omega- external resistance- What will be the potential difference across the terminals of the cell -

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Standard XII

Physics

Parallel and Series Combination of Capacitors

A cell of emf...

Question

A cell of emf 2V and internal resistance 0.1 Ω is connected to a 3.9 Ω external resistance. What will be the potential difference across the terminals of the cell ?

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Solution

Internal resistance of the battery and the external resistance are connected in series.
R = 3.9 + 0.1 = 4.0 ohm
Current in the circuit = E/R = 2/4 = 0.5 A

terminal voltage of the battery = E - Ir
= 2 - 0.5*0.1 = 1.95 volt.

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A cell of emf 2V and internal resistance 0.1 Ω is connected to a 3.9 Ω external resistance. What will be the potential difference across the terminals of the cell ?

Internal resistance of the battery and the external resistance are connected in series. R = 3.9 + 0.1 = 4.0 ohm Current in the circuit = E/R = 2/4 = 0.5 A terminal voltage of the battery = E - Ir = 2 - 0.5*0.1 = 1.95 volt.

Internal resistance of the battery and the external resistance are connected in series.
R = 3.9 + 0.1 = 4.0 ohm
Current in the circuit = E/R = 2/4 = 0.5 A

terminal voltage of the battery = E - Ir
= 2 - 0.5*0.1 = 1.95 volt.