Answer to Question #144998 in Electric Circuits for Edward Lawerence

Question

The load was connected to the external battery circuit. At a current of 3 A, 18 W of power is
released in the external circuit. Find the load resistance and EMF of the battery if the internal resistance of
the battery is 0.5 .

Answer 1

Explanations & Calculations

• The total consumption of power ($\small P$ )of the load ($\small R$ ) is given by, $\small P = Vi=i^2R=\large \frac{V^2}{R}$.
• Therefore, using the appropriate equation ($\small P=i^2R\,,P=Vi$) with the given data (voltage & current), the load resistance & the voltage applied ($\small V_{applied}$ ) to the external circuit could be calculated.
• Therefore,

$\qquad\qquad \begin{aligned} \small 18W&= \small (3A)^2\times R_{load}\\ \small R_{load}&= \small \bold{2\Omega}\\ \\ \small V_{applied} &= \small \frac{P}{i}= \frac{18W}{3A}=6V \end{aligned}$

• In fact, all the power sources have some internal resistance which drops some voltage across them resulting a lower applied voltage than the EMF of the source.
• When power is given out of the source to an external circuit, the applied/available voltage is given as $\small V_{applied}=E-i_{applied}R_{internal}$.
• 6V is the applied voltage & the 3A is solely conducted through the source thus the applied current.
• Therefore, the EMF is,

$\qquad\qquad \begin{aligned} \small EMF &= V_{applied}+iR\\ &= \small 6V+(3A\times 0.5\Omega)\\ &= \small \bold{7.5V} \end{aligned}$