You have constructed an RC circuit to the right. A power...

You have constructed an RC circuit to the right. A power supply, resistor, and capacitor are all connected in series with $epsilon = 7$ V, R = 300$Omega$, and C = 15 $mu$F respectively. When t = 0, the capacitor is completely uncharged. A. What is the time constant for the circuit? B. What is the voltage across the capacitor at t = 25 ns? C. How much time it will take to charge the capacitor to 70%?

Transcript

00:01 In this problem we have an rc circuit and initially the capacitor is completely uncharged so for the first part we have to find out the time constant time constant is equal to the product of the resistance and capacitance in an rc circuit now the resistance is 300 oms and the capacitance is 15 micro fared now micro farad is equal to 10 to the power minus 6 ferret so we get time constant 4 .5 times 10 to the power minus 3 seconds this is the time constant of the rc circuit now for the second part we have to find out the voltage across the capacitor at the given time now for an rc circuit the voltage across capacitor at time t is given by b not which is the initial voltage sorry the maximum voltage that is called this v max maximum voltage times 1 minus a to the power minus t upon and constant now the maximum voltage across the capacitor will be the voltage of battery when the capacitor is fully charged so when capacitor is fully charged its potential will be 7 volts equal to the potential of battery and it will be the maximum voltage across the capacitor now time t is 25 nanoseconds which is 25 times 10 to the power minus 9 seconds upon time constant which is 4 .5 times 10 to the power minus 3 so upon solving this we get the sign of the power of exponential is negative so this will be negative and this gives us 3 .89 times 10 to the power minus 5 volts this is the potential at 25 nanoseconds now we have to find out the time taken for the capacitor to charge to 70 percent so the charge at times t is equal to maximum charge when it is fully charged times one minus e to the power minus t upon time constant now it is charged to 70 % which means the charge at time t is 70 % of the maximum charge so 70 upon 100 times q max equal to q max times 1 minus e to the power minus t upon time constant so q max cancels out and we get 0 .7 equal to 1 minus a to the power minus t upon time constant from this we get 1 minus 0 .7 equal to 0 .3 taking natural log both sides we get log e to the power minus t upon time constant equal to natural log 0 .3 so on the left hand side we get minus t upon time constant equal to log point three so time t is equal to minus time constant times log point three time constant is 4 .5 times 10 to the power minus 3 seconds times log 0 .3 so this gives us time equal to 5 .42 times 10 to the power minus 3 seconds this is a time when the capacitor is charged to 70%...

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