Question #129386

Find out the time at which the potential across the capacitor is equal to that across the
resistor, when switch S is closed at t=0, the capacitor is charged through the resistor.
Consider the value of capacitance of a capacitor= 12mF and resistance of resistor is 25ohm.

Expert's answer

The voltage across a capacitor during charging can be describe with the following formula:


UC=U0(1−e−tRC)U_C = U_0(1 - e^{-\frac{t}{RC}})

where U0U_0 is the voltage of the battery, tt is time, C=12×10−3FC = 12\times 10^{-3}F is the capacitance of the capacitor and R=25ΩR = 25\Omega is the resistance of the resistor.

On the other hand, the voltage across the resistor is:


UR=U0−UC=U0−U0(1−e−tRC)=U0e−tRCU_R = U_0 - U_C = U_0 - U_0(1 - e^{-\frac{t}{RC}}) = U_0e^{-\frac{t}{RC}}

The required condition is:


UC=URU_C = U_R\\

U0(1−e−tRC)=U0e−tRCe−tRC=12U_0(1 - e^{-\frac{t}{RC}}) = U_0e^{-\frac{t}{RC}}\\ e^{-\frac{t}{RC}} = \dfrac12

Expressing the time from the last equation, get:


t=−RCln⁡12=−25⋅12⋅10−3ln⁡12≈0.21st = -RC\ln\dfrac12 = -25\cdot 12\cdot 10^{-3}\ln \dfrac12\approx 0.21s

Answer. 0.21 s.


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