### Video Transcript

A circuit contains a D cell battery, a switch, a 20-ohm resistor, and four 20- millifarad capacitors connected in series. What is the equivalent capacitance of the circuit? What is the RC time constant? How long before the current decreases to 50 percent of the initial value once the switch is closed?

Weโre told the resistance of the circuit is 20 ohms. Weโll call that capital ๐
. Also in the circuit are four 20-millifarad capacitors. Weโll call the value of each one individually capital ๐ถ. In this three-part problem, we wanna solve first for the equivalent capacitance of the circuit. Weโll call that ๐ถ sub ๐. We next wanna solve for the RC time constant, which weโll call ๐. And finally, we wanna solve for a time value, how long before the current decreases to half its original value. Weโll call that time ๐ก sub one-half.

Letโs start by drawing a sketch of this circuit. In this series circuit, we have our D cell power supply, weโve called it ๐, our resistor ๐
of given value 20 ohms, and four capacitors are again arranged in series followed by a switch which can close to complete the circuit. We wanna solve first for the total equivalent capacitance of the circuit. And to do that, letโs recall the rule for adding capacitors in series.

One over the total equivalent capacitance, ๐ถ sub ๐, is equal to one over the first capacitorโs value plus one over the second capacitorโs value up to one over the last capacitorโs value. In our case, we have four capacitors in our circuit. And they all have the same capacitance value, ๐ถ, which means we can write their sum as four divided by ๐ถ. When we cross multiply this expression to solve for ๐ถ sub ๐, we find itโs equal to ๐ถ, our capacitance value of 20 millifarads, divided by four. When we plug in that value for ๐ถ and calculate this fraction on our calculator, we find that ๐ถ sub ๐ is 5.0 millifarads.

Thatโs the equivalent capacitance of the circuit. Next we wanna to solve for ๐, the time constant in this RC circuit. In a circuit of this type, the time constant is equal to the resistance in the circuit multiplied by its capacitance. Applying this relationship to our scenario, we would write that ๐ equals ๐
times ๐ถ sub ๐, the equivalent capacitance in the whole circuit. Weโve been given the value of ๐
. And weโve solved for ๐ถ sub ๐. So we can now plug in to solve for ๐.

When we do and calculate this result, we find the ๐ is 100 milliseconds. Thatโs this circuitโs time constant. Finally, we want to solve for ๐ก sub one-half. After the switch and the circuit is closed, current will start of flow in the circuit. At first it will have its maximum value. But as the capacitors soak up charge, the amount of current left in the circuit will decrease with time. ๐ก sub one-half is the time it would take for the current in the circuit to decrease to half its maximum value.

To solve for ๐ก sub one- half, letโs recall the mathematical relationship for current in an RC circuit. Current as a function of time is equal to the initial potential difference divided by the total circuit resistance times ๐ to the negative ๐ก over ๐, the time constant. This equation shows us that when ๐ก is equal to zero โ that is, at the initial moment the circuit is connectedโthe maximum current in the circuit is equal to ๐ over ๐
. So we can rewrite this expression so that current as a function of ๐ก is equal to the maximum current in the system multiplied by ๐ to the negative ๐ก over ๐.

We want to solve for that value of time such that when ๐ก is equal to ๐ก one-half and that particular value in time, the current is equal to half the maximum current, ๐ผ sub max over two. When we look at this equation, we see that ๐ผ sub max appears on both sides. So we can cancel it out. And our final result will be independent of it. So one-half is equal to ๐ to the negative ๐ก sub one-half over ๐. If we take the natural logarithm of both sides, then the expression on the right-hand side of our equation simplifies to negative ๐ก sub one-half over ๐.

If we multiply both sides then by negative ๐, rearranging so that ๐ก sub one-half is on the left, we see itโs equal to negative ๐ times the natural log of one-half. Weโve solved for ๐, the time constant, in an earlier part of the problem. So we can now plug in that value to solve for ๐ก sub one-half. When we make that substitution and calculate this expression, we find that, to two significant figures, ๐ก sub one-half is 69 milliseconds. Thatโs how long it would take the current in this circuit to decrease to one-half its initial maximum value.