Question

Write the set of equations that determines the three currents in the circuit shown in the figure. (Assume that the capacitor is initially uncharged.) V + (V)e^-2, 990t s^-1 - 85.0 V = 0 85.0 V - [V + (V)e^-2, 990t s^-1] = 0 V + (V)e^-2, 990 t s^-1 +170 V = 0 (-0.521 A) e^-2, 990 t s^-1 = (-1.30 A)e^-2, 990 t s^-1 + (A)e^-2, 990 t s

Answer 1

In an RC (series) circuit, KVL can be given by: V – I R –V_C = 0.

But V_C = V e(^–t/RC)

Using this, the equations in different loops shown here can be:

i₂ × 32 + 85 e(^–t/RC) – 85 = 0 .....(i)

85 – (i₃ × 21.5 + 85 e(^–t/RC) ) = 0 .....(ii)        

170 – i₂ × 32 – i₃ × 21.5 = 0 ....(iii)

Also , i₂ = i₁ + i₃ ....(iv)

For both (i) and (ii), R is the effective resistance. Here, 32 $\Omega$ and 21.5 $\Omega$are in parallel combination which is in series to the capacitor. So R is to be calculated as the effective resistance for the parallel combination of 32 $\Omega$ and 21.5 $\Omega$.

The value of C is given. Equations can be easily formulated.