The figure shows a branch of an electrical circuit. At a given...

Applying Kirchhoff's Voltage Law (KVL) from point Q to point P, we can write the equation for the potential difference:

$V_Q - I R - L \dfrac{dI}{dt} + E = V_P$

Given the parameters in the problem:
$\bullet$ Point Q is grounded, so $V_Q = 0 \text{ V}$.
$\bullet$ The current $I$ flowing from Q to P is $3 \text{ A}$.
$\bullet$ The current is decreasing at a rate of $4 \text{ A/s}$, so the rate of change of current is $\dfrac{dI}{dt} = -4 \text{ A/s}$.
$\bullet$ The resistance $R = 4 \ \Omega$.
$\bullet$ The inductance $L = 0.5 \text{ H}$.
$\bullet$ The battery has an EMF $E = 6 \text{ V}$. Traversing from Q to P, we move from the negative (shorter, thicker line) to the positive (longer, thinner line) terminal, which means there is a potential gain of $+6 \text{ V}$.

Substituting these values into the KVL equation:

$0 - (3 \text{ A})(4 \ \Omega) - (0.5 \text{ H})(-4 \text{ A/s}) + 6 \text{ V} = V_P$

$0 - 12 \text{ V} + 2 \text{ V} + 6 \text{ V} = V_P$

$V_P = -4 \text{ V}$

The electric potential at point P is $-4 \text{ V}$.

Answer: $-4 \text{ V}$