Irodov 3.192: Where a Battery's Power Goes – Amit & Deepali Sinha Foundation

Problem Irodov 3.192

A storage battery of emf $\mathcal{E} = 2.6$ V drives a current $I = 1.0$ A through an external resistance, and the potential difference between its terminals is then $V = 2.0$ V. Find the heat generated per unit time inside the battery, and the power developed inside the battery by the electric forces.

The battery as its emf E with its internal resistance r, between its terminals a (+) and b (-), driving the current I through the external load R. — The battery as its emf $\mathcal{E}$ with its internal resistance $r$ , between its terminals $a$ ( $+$ ) and $b$ ( $-$ ), driving the current $I$ through the external load $R$ .

Given

$\mathcal{E} = 2.6$ V
$I = 1.0$ A
terminal voltage $V = 2.0$ V

Find

inside the battery: the heat per unit time

Q

· the power

P

of the electric forces

Solution

Step 1The internal resistance from the terminal voltage

Inside the battery the current flows from the negative terminal $b$ to the positive terminal $a$ , through the emf and the internal resistance $r$ . Going that way, the emf raises the potential by $\mathcal{E}$ and the resistance lowers it by $Ir$ , so the terminal voltage is

V = \varphi_a - \varphi_b = \mathcal{E} - Ir

r = \frac{\mathcal{E} - V}{I} = 0.60~\Omega

(1)

Step 2The heat generated in the battery

Joule’s law for the internal resistance, with (1):

Q = I^2r = I^2\,\frac{\mathcal{E} - V}{I}

Q = I(\mathcal{E} - V) = 0.6\text{ W}

(2)

Step 3The power of the electric forces inside the battery

Each charge $dq$ that crosses the battery goes from $b$ to $a$ , and the electric forces do on it the work $dq\,(\varphi_b - \varphi_a)$ . Dividing by $dt$ , with $dq/dt = I$ :

P = I\,(\varphi_b - \varphi_a)

P = -IV = -2.0\text{ W}

(3)

The negative sign means the electric forces take $2.0$ W from the carriers inside the battery; that is the power the battery hands on to the external circuit, where the same field does positive work $IV$ .

Check

The energy balance inside the battery. The extraneous forces do the work $\mathcal{E}\,dq$ on each charge, a power $\mathcal{E}I = 2.6$ W. Together with the electric forces they supply the heat:

\mathcal{E}I + P = 2.6\text{ W} - 2.0\text{ W} = 0.6\text{ W} = Q

as (2) and (3) require. Dimensions: volts times amperes are watts in both answers.

Remark

The terminal voltage measures how much of the emf survives the internal drop: here $V/\mathcal{E} = 0.77$ , so $77$ per cent of the power the chemistry develops reaches the load and $23$ per cent heats the battery. On charging, with the current reversed through the battery, $\varphi_a - \varphi_b$ exceeds $\mathcal{E}$ , the electric forces do positive work inside, and the heat is still $I^2r$ , now paid by the charger.