Why do different resistor values dissipate power at different rates in RL and RC circuits?
but I think the current caused by this voltage will be proportional
  to the resistance values and so will the power dissipated by this


Let's work it out and see.  We assume that the source is disconnected at t=0.

For RL and RC circuits with initial energy, i.e., there is an initial current through the inductor or an initial voltage across the capacitor, the current is given by:

\$i(t) = i_0 \ e^{-t/\tau} \$

For the RL circuit:

\$\tau = L/R, \ i_0 = i_L(0)  \$

For the RC circuit:

\$\tau = RC, \ i_0 = v_C(0) / R  \$

Now, let's calculate the power associated with the resistor R:

\$p_R(t) = i^2(t) R = i^2_0\ e^{-2t/\tau} \$

For the RL circuit:

\$p_R(t) = i^2_L(0)\ e^{-2t/\tau}\ R\$

As we expect for the RL circuit, the power is proportional to the resistance R.

For the RC circuit:

\$p_R(t) = \dfrac{v^2_C(0)}{R^2}\ e^{-2t/\tau}\ R = \dfrac{v^2_C(0)}{R}\ e^{-2t/\tau}\$

So, for the RC circuit, the power is inversely proportional to the resistance R.

How to intuit this without working through the math?  Note that the larger the power, the sooner the initial energy is dissipated, i.e., the smaller the time constant.

Now, note the formula for the time constant.  For the RL circuit, a larger R gives a smaller time constant while, for the RC circuit, a smaller R gives a smaller time constant.