Question

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?

Answer 1

The voltage is
`V =   0.993V_b`

Step-by-step explanation:

From the question we are told that

The time that has passed is
`t = (\tau)/(2)`

Here
`\tau` is know as the time constant

The voltage of the power source is
`V_b`

Generally the voltage equation for charging a capacitor is mathematically represented as

`V =  V_b  [1 - e^{- (t)/(\tau) }]`

=>
`V =  V_b  [1 - e^{- ((\tau)/(2))/(\tau) }]`

=>
`V =  V_b  [1 - e^{- (\tau)/(2\tau) }]`

=>
`V =  V_b  [1 - e^{- (1)/(2) }]`

=>
`V =   0.993V_b`