Current Electricity: Application Problems (1)

Question

A battery of EMF $\varepsilon = 12$ V and internal resistance $r = 1\ \Omega$ is connected to an external resistor $R = 5\ \Omega$. Find the current in the circuit, the terminal voltage, and the power dissipated in $R$.

A clean Class 10 / Class 12 question. The trap is small but real, and most students get it on the first try only after they’ve been burned once.

Solution — Step by Step

Why This Works

A real battery is modelled as an ideal EMF $\varepsilon$ in series with a small internal resistance $r$. When current flows, some of the EMF is “used up” inside the battery itself — that’s why terminal voltage drops below $\varepsilon$ during discharge.

When the circuit is open ($I = 0$), the terminal voltage equals $\varepsilon$. As current increases, terminal voltage falls linearly: $V = \varepsilon - Ir$. This is why your phone “shows full battery” until you start using it.

Alternative Method

Use power conservation. $P_{total} = \varepsilon I = 12 \cdot 2 = 24$ W. $P_r = I^2 r = 4$ W. So $P_R = 24 - 4 = 20$ W. ✓

Common Mistake

Students often write $I = \varepsilon/R = 12/5 = 2.4$ A, ignoring the internal resistance. Whenever the question gives $r$, you must include it in the loop equation. If a battery is described as “ideal” or “internal resistance negligible,” only then can you skip $r$.

Another trap: the “terminal voltage” reading is not the EMF unless current is zero. During discharge, $V < \varepsilon$. During charging (current pushed against the EMF), $V > \varepsilon$.