Answer to Question #185045 in Electric Circuits for Niall Rio

Question

When the switch is open, the potential difference across the terminals of a battery is 20.0 V. When a current of 2.0 A flows, the potential difference across the battery drops to 16 V. (a) If this battery consists of two identical cells in series, what is the electromotive force of each cell? (b) If this battery consists of two identical cells in series, what is the internal resistance of each cell? (c) If this battery consists of two identical cells in parallel, what is the electromotive force of each cell? (d) If this battery consists of two identical cells in parallel, what is the internal resistance of each cell?

Answer 1

Explanations & Calculations

• When a battery with internal resistance operates, the potential difference across it is given by

$\qquad\qquad \begin{aligned} \small V&=\small E\pm ir\begin{cases} \small E+ir: \text{during being charged}\\ \small E-ir: \text{during normal operation} \end{cases} \end{aligned}$

• Therefore, from the given data,

$\qquad\qquad \begin{aligned} \small 16\,V&=\small20\,V-2A\times r\\ \small r&=\small 2\, \Omega \end{aligned}$

the total equivalent resistance of the battery pack is $\small 2\Omega$ and the total electromotive force is $\small 20\,V$.

• When the batteries are in series connection, both their electromotive forces & the internal resistances just add up. Therefore,

$\qquad\qquad \begin{aligned} \small 20V&=\small E_1+E_2\\ \small E_1&=\small E_2\\ \small 2E&=\small 20\\ \small E&=\small \bold{10V}\\\\ \small 2\Omega &=\small r_1+r_2\\ \small r_1&=\small r_2\\ \small 2r&=\small 2\\ \small r&=\small \bold{1 \Omega} \end{aligned}$

• When they are connected parallel their equivalent resistance reduces according to the general rule, therefore,

$\qquad\qquad \begin{aligned} \small \frac{1}{2\Omega}&=\small \frac{1}{r_1}+\frac{1}{r_2}\\ \small r_1 &=\small r_2\\ \small \frac{2}{r}&=\small \frac{1}{2}\\ \small r&=\small \bold{4\,\Omega} \end{aligned}$

• And their electromotive forces are given according to the equation $\frac{E_e}{r_e}=\frac{E_1}{r_1}+\frac{E_2}{r_2}$
• Therefore,

$\qquad\qquad \begin{aligned} \small \frac{20V}{2\Omega}&=\small \frac{E_1}{4\Omega}+\frac{E_2}{4\Omega}\\ \small E_1&=\small E_2\\ \small 10&=\small \frac{2E}{4}\\ \small E&=\small \bold{20\,V} \end{aligned}$