Consider the circuit displayed in figure (a). Determine the current...

In both circuits, the total resistance is the sum of the series resistors:

$R_{\text{eq}} = 10\text{ }\Omega + 5\text{ }\Omega = 15\text{ }\Omega$

For the circuit in figure (a), the two batteries are connected such that their polarities aid each other (both have their positive terminals on the left). The net electromotive force (EMF) is:

$E_{\text{net, a}} = 12\text{ V} + 6\text{ V} = 18\text{ V}$

The current in circuit (a) is:

$I_{\text{a}} = \dfrac{E_{\text{net, a}}}{R_{\text{eq}}} = \dfrac{18}{15} = 1.2\text{ A}$

The potential drop across the $5\text{ }\Omega$ resistor is:

$V_{5\Omega} = I_{\text{a}} \times 5 = 1.2 \times 5 = 6\text{ V}$

The potential drop across the $10\text{ }\Omega$ resistor is:

$V_{10\Omega} = I_{\text{a}} \times 10 = 1.2 \times 10 = 12\text{ V}$

For the circuit in figure (b), the batteries are connected in opposition (the $12\text{ V}$ battery has its positive terminal on the left, while the $6\text{ V}$ battery has its positive terminal on the right). The net EMF is:

$E_{\text{net, b}} = 12\text{ V} - 6\text{ V} = 6\text{ V}$

The current in circuit (b) is:

$I_{\text{b}} = \dfrac{E_{\text{net, b}}}{R_{\text{eq}}} = \dfrac{6}{15} = 0.4\text{ A}$

The potential drops in circuit (b) are:

$V_{5\Omega} = 0.4 \times 5 = 2\text{ V}$

$V_{10\Omega} = 0.4 \times 10 = 4\text{ V}$

Since the current and potential drops in figure (b) are different from those in figure (a), the values are not the same.

Answer: $1.2\text{ A}$, $6\text{ V}$, and $12\text{ V}$ respectively; the values are different for figure (b)