Q.A battery of emf 12 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, a] Calculate resistance of resistor. b] Calculate terminal voltage of the battery when the circuit is closed.

An EMF DEVICE (a cell, battery, solar cell, fuel cell, or generator) maintains a potential difference across its own terminals by doing WORK on the charge carriers passing through it, driving them from low to high potential energy internally using some non-electrostatic energy source (chemical energy in an ordinary cell, absorbed photon energy in a solar cell). Its emf, $\varepsilon=dw/dq$, is the work done per unit charge, with SI unit joule/coulomb (equivalently the volt).

An IDEAL emf device has no internal resistance, so its emf always equals its terminal voltage exactly. A REAL device has some internal resistance $r$, so when current $I$ flows, the terminal voltage falls short of the full emf: $V=\varepsilon-Ir$. Combined with Ohm's law applied to an external resistor $R$, this gives the current actually delivered to the circuit, $I=\varepsilon/(R+r)$. The MAXIMUM current the device can ever deliver occurs in the extreme case $R=0$ (short circuit), $I_{max}=\varepsilon/r$ -- a value set entirely by the device's own internal resistance, and what fixes a real cell or battery's maximum current rating (this is also why an aged battery, whose internal resistance has grown large, can no longer deliver the large currents it once could).