Skip to main content
Electrical Engineering

Return to Answer

added 67 characters in body
Source Link
Chu
Chu
  • 7.7k
  • 2
  • 15
  • 16

The switch is closed at \$\small t=0\$, and the capacitor voltage for \$\small t \ge 0\$ is: \$\small v_c(t)=40-\frac{1}{C}\int i\: dt\$, where \$\small i\$ is the capacitor discharge current.

Now use KCL at the node pointed to by the blue arrow to form the required differential equation, and solve (using the integrating factor method, for example). The initial capacitor current is \$\small i(0)=\frac{10}{22} \:A\$

The switch is closed at \$\small t=0\$, and the capacitor voltage for \$\small t \ge 0\$ is: \$\small v_c(t)=40-\frac{1}{C}\int i\: dt\$, where \$\small i\$ is the capacitor discharge current.

Now use KCL at the node pointed to by the blue arrow to form the required differential equation, and solve (using the integrating factor method, for example).

The switch is closed at \$\small t=0\$, and the capacitor voltage for \$\small t \ge 0\$ is: \$\small v_c(t)=40-\frac{1}{C}\int i\: dt\$, where \$\small i\$ is the capacitor discharge current.

Now use KCL at the node pointed to by the blue arrow to form the required differential equation, and solve (using the integrating factor method, for example). The initial capacitor current is \$\small i(0)=\frac{10}{22} \:A\$

Source Link
Chu
Chu
  • 7.7k
  • 2
  • 15
  • 16

The switch is closed at \$\small t=0\$, and the capacitor voltage for \$\small t \ge 0\$ is: \$\small v_c(t)=40-\frac{1}{C}\int i\: dt\$, where \$\small i\$ is the capacitor discharge current.

Now use KCL at the node pointed to by the blue arrow to form the required differential equation, and solve (using the integrating factor method, for example).