A leaky 100 microfarad capacitor is charged to a potental difference

[islandguy]

A leaky 100 microfarad capacitor is cahrged to a potental difference of 50V. IT leaks as though a Kilo ohm is connected across it.

a) Find the initial discharge current.
b) If a child can feel an electric shock when the voltage of a source is higherthan 20 V. after what time can the child touch the terminals of the capacitor, without feeling any electric shock.



Please HElp Me
very Urgent
Thanks in advance

 

[chroot]

The current through a resistance can be found with Ohm's law: V = IR.

The current is also the rate of passage of charge (I = dQ/dt), and the voltage on a capacitor is related to its charge by Q = CV. You can easily solve for the voltage on the capacitor at any time.

- Warren

 

[M98Ranger]

a) To find the initial discharge current, we can use the formula I = C(dV/dt), where I is the current, C is the capacitance, and dV/dt is the rate of change of voltage. In this case, C = 100 microfarads and dV/dt = -50V (since the capacitor is discharging). Therefore, the initial discharge current would be -50V*(100 microfarads) = -0.005 amps.

b) To calculate the time it takes for the voltage to drop below 20V, we can use the formula V = V0*e^(-t/RC), where V0 is the initial voltage, t is the time, R is the resistance, and C is the capacitance. In this case, V0 = 50V, R = 1000 ohms, and C = 100 microfarads. Plugging these values in, we get 20V = 50V*e^(-t/(1000 ohms * 100 microfarads)). Solving for t, we get t = 0.138 seconds. Therefore, after 0.138 seconds, the child can touch the capacitor's terminals without feeling an electric shock.