Physics - Current Electricity Question with Solution | TestHub

From the given circuit diagram, the two batteries are connected in opposition as their positive terminals face each other.

The net electromotive force (EMF) in the circuit is:
$E_{\text{net}}=8\text{ V}-4\text{ V}=4\text{ V}$

The total resistance of the series circuit is the sum of all resistances:
$R_{\text{total}}=R+r_1+r_2$
$R_{\text{total}}=9+1+2=12\Omega$

The current $I$ flowing through the circuit is:
$I=\frac{E_{\text{net}}}{R_{\text{total}}}=\frac{4}{12}=\frac{1}{3}\text{ A}$

The potential difference between points P and Q is the voltage across the external resistor $R=9\Omega$:
$V_{PQ}=I\times R$
$V_{PQ}=\frac{1}{3}\times 9=3\text{ V}$

Answer: $\frac{1}{3}\mathrm{A}$ and 3 V