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?
We’re told the resistance of the circuit is 20 ohms. We’ll call that capital 𝑅. Also in the circuit are four 20-millifarad capacitors. We’ll call the value of each one individually capital 𝐶. In this three-part problem, we wanna solve first for the equivalent capacitance of the circuit. We’ll call that 𝐶 sub 𝑇. We next wanna solve for the RC time constant, which we’ll call 𝜏. And finally, we wanna solve for a time value, how long before the current decreases to half its original value. We’ll call that time 𝑡 sub one-half.
Let’s start by drawing a sketch of this circuit. In this series circuit, we have our D cell power supply, we’ve called it 𝑉, our resistor 𝑅 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, let’s recall the rule for adding capacitors in series.
One over the total equivalent capacitance, 𝐶 sub 𝑇, is equal to one over the first capacitor’s value plus one over the second capacitor’s value up to one over the last capacitor’s value. In our case, we have four capacitors in our circuit. And they all have the same capacitance value, 𝐶, which means we can write their sum as four divided by 𝐶. When we cross multiply this expression to solve for 𝐶 sub 𝑇, we find it’s equal to 𝐶, our capacitance value of 20 millifarads, divided by four. When we plug in that value for 𝐶 and calculate this fraction on our calculator, we find that 𝐶 sub 𝑇 is 5.0 millifarads.
That’s the equivalent capacitance of the circuit. Next we wanna to solve for 𝜏, 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 𝜏 equals 𝑅 times 𝐶 sub 𝑇, the equivalent capacitance in the whole circuit. We’ve been given the value of 𝑅. And we’ve solved for 𝐶 sub 𝑇. So we can now plug in to solve for 𝜏.
When we do and calculate this result, we find the 𝜏 is 100 milliseconds. That’s this circuit’s time constant. Finally, we want to solve for 𝑡 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. 𝑡 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 𝑡 sub one- half, let’s 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 𝑒 to the negative 𝑡 over 𝜏, the time constant. This equation shows us that when 𝑡 is equal to zero — that is, at the initial moment the circuit is connected—the maximum current in the circuit is equal to 𝑉 over 𝑅. So we can rewrite this expression so that current as a function of 𝑡 is equal to the maximum current in the system multiplied by 𝑒 to the negative 𝑡 over 𝜏.
We want to solve for that value of time such that when 𝑡 is equal to 𝑡 one-half and that particular value in time, the current is equal to half the maximum current, 𝐼 sub max over two. When we look at this equation, we see that 𝐼 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 𝑒 to the negative 𝑡 sub one-half over 𝜏. If we take the natural logarithm of both sides, then the expression on the right-hand side of our equation simplifies to negative 𝑡 sub one-half over 𝜏.
If we multiply both sides then by negative 𝜏, rearranging so that 𝑡 sub one-half is on the left, we see it’s equal to negative 𝜏 times the natural log of one-half. We’ve solved for 𝜏, the time constant, in an earlier part of the problem. So we can now plug in that value to solve for 𝑡 sub one-half. When we make that substitution and calculate this expression, we find that, to two significant figures, 𝑡 sub one-half is 69 milliseconds. That’s how long it would take the current in this circuit to decrease to one-half its initial maximum value.