Question

The capacitors are initially uncharged, In a certain time the capacitor of capacitance $2\mu F$ gets a charge of $20\mu C$. In that time interval find the heat produced by each resistor individually

Answer 1

$q_2=20\mu C$

$\therefore q_1=10\mu C$ (as they are in parallel)

Energy stored at this instant,

$U=\frac{1}{2}\frac{q_1^2}{C_1}+\frac{1}{2}\frac{q_2^2}{C_2}$

$=\frac{1}{2}\times \frac{({10}^{-5}{)}^2}{{10}^{-6}}+\frac{1}{2}\times \frac{(2\times {10}^{-5}{)}^2}{2\times {10}^{-6}}$

$=1.5\times {10}^{-4}J$

$=0.15mJ$

In charging of a capacitor 50% of the energy is stored and rest 50% is dissipated in the form of heat Therefore 0,15 mJ will be dissipated in the form of heat across an the resistors. In series in direct ratio of resistance $(H=i^{2}Rt)$ and in parallel in inverse ratio of resistance.

$\therefore H_2=0.075mJ,H_3=0.05mJ$

and $H_6=0.025mJ$