Chapter 10, Problem 10.1CYU

Check Your Understanding If you place a wire directly across the two terminal of a battery, effectively shorting out the terminals, the battery will begin to get hot. Wiry do you suppose this happens?

To determine

The reason for the battery getting heated while shorting out its two terminals.

Explanation of Solution

Introduction:

If a battery with internal resistor r is connected to a load resistor R, the load resistor is in series with the internal resistor. The equivalent resistance of the circuit becomes $\left(r+R\right)$.

Applying Ohm's law in the circuit

The magnitude of the current flowing through the resistor is, $I=\frac{\epsilon }{r+R}$, and

Power consumed by a resistor, $P=\frac{V^2}{R}=I^{2}R$

Short circuit of battery occurs, if one connectsthe positive terminal of the battery to the negative terminal with highly conducting wire. Due to low load resistance, a huge amount of current starts to flow.

One can assume resistance of the connecting wire is zero, then the magnitude of the current flow through the circuit is

  $\begin{array}{l}I=\frac{\epsilon }{R+r}\\ =\frac{\epsilon }{0+r}\\ =\frac{\epsilon }{r}\end{array}$

The power dissipated in internal resistance,

  $\begin{array}{l}P=I^{2}r\\ ={\left(\frac{\epsilon }{r}\right)}^{2}r\\ =\frac{\epsilon }{r^2}\end{array}$

Conclusion:

Power dissipated in the internal resistance will generate heat, and the battery begins to get hot.