Answer
$1800\;\rm C$
Work Step by Step
e know, according to Ohm's law, that
$$ V_{ B}=IR$$
Now we need to find the initial current at $t=0$
$$I_0=\dfrac{ (V_B)_0}{R}$$
where $(\Delta V_B)_0=1.5$ at $t=0$
$$I_0=\dfrac{ 1.5}{3}=\bf 0.5\;\rm A$$
and at $t=2$ h, the current is
$$I_f=\dfrac{ (V_B)_f}{R}=\dfrac{0}{3}=\bf 0\;\rm A$$
Now we can draw the $I$ versus $t$ graph, as we see below.
Recalling that
$$I=\dfrac{dQ}{dt}$$
So,
$$dQ=Idt$$
Taking the integral,
$$\int_0^QdQ=\int_0^t Idt$$
$$ Q=\int_0^t Idt$$
where $\int_0^t Idt$ is the area under the $I$ versus $t$ graph.
Thus,
$$ Q=\frac{1}{2}\Delta t \Delta I=\frac{1}{2}(2\times 3600)(0.5)$$
$$ Q=\color{red}{\bf 1800}\;\rm C$$
