Chapter 6.3, Problem 1aT

A capacitor is connected to a battery, bulb, and switch as shown. Assume that the switch has been closed for an extended period of time.

1. Predict whether the brightness of the bulb is the same as, greater than, or less than the brightness of a single bulb connected to a battery. Explain.

2. Predict how the potential difference across the battery to the potential differences across the capacitor plates and to the potential difference across the bulb. Explain.

3. Briefly describe the distribution of charge, if any, on the capacitor plates.

Recall the relationship between the charge on a capacitor and the potential difference across the capacitor. Use this relationship to describe how you could use a voltmeter to determine the charge on a capacitor.

4. Obtain the circuit and a voltmeter. Check your predictions for parts 1 and 2.

To determine

To Explain:Whether the brightness of the bulb is the same as, greater than or less than the brightness of a single bulb connected to a battery.

Answer to Problem 1aT

Brightness of the bulb changes with the time in RC circuit but it is constant in case of battery only.

Explanation of Solution

Introduction:

Ohm’s Law: The current in the circuit is directly proportional to the potential difference and the constant of proportionality is known as resistance R.

  $V=I\times R$

The potential difference across the capacitor is given as: $V=\frac{Q}{C}$

Where, $Q$ is the charge on the capacitor and $C$ is the capacitance of the capacitor.

A RC circuit with a battery connected to the capacitor through a Bulb with a resistance of $R$ shown in Figure 1. The capacitor starts charging and it will charge till the potential difference of the capacitor equals to the voltage of the battery. Initially, as there is no charge on capacitor, it behaves as a short circuit, hence maximum current is drawn from the battery in the circuit using Ohm’s law $V_{battery}=I\times R$ , R is the resistance of the bulb. As the time passes, charge starts accumulating on capacitor and there is an increase in potential difference on the capacitor.

  

Figure 1: A RC circuit with a battery, bulb and a capacitor

After a long time, the circuit will behave as an open circuit and Therefore, current in the circuit drops to zero with the time till the capacitor voltage equals to the battery voltage.During initial moment, the brightness of the bulb in this RC is equivalent to the single bulb with the battery.But as the time passes, current starts dropping, hence, brightness decreases till the circuit becomes open due to the charging of the capacitor.

Conclusion:

Brightness of the bulb changes with the time in RC circuit but it is constant in case of battery only.

To determine

To Compare: The potential difference across capacitor, bulb and battery.

Answer to Problem 1aT

The voltage difference between the terminals of the battery is $V_{battery}$ which is constant and the voltage difference across the capacitor $V_c\left(t\right)=\frac{Q\left(t\right)}{C}$ and across the bulb is $V_{Bulb}\left(t\right)=I\left(t\right)\times R$ .

Explanation of Solution

Introduction:

Ohm’s Law: The current in the circuit is directly proportional to the potential difference and the constant of proportionality is known as resistance R.

  $V=I\times R$

The potential difference across the capacitor is given as:

  $V=\frac{Q}{C}$

Where $Q$ is the charge on the capacitor and $C$ is the capacitance of the capacitor.

A RC circuit with a battery connected to the capacitor through a Bulb with a resistance of $R$ shown in Figure 2. The voltage difference across the capacitor and across the bulb is time dependent.

  

Figure 2: A RC circuit with a battery, bulb and a capacitor

The voltage difference between the terminals of the battery is $V_{Battery}$ ,the voltage difference across the capacitor $V_c\left(t\right)=\frac{Q\left(t\right)}{C}$ and across the bulb is $V_{Bulb}\left(t\right)=I\left(t\right)\times R$ .

Hence, the voltage through the loop can be written as:

  $V_{Battery}-V_{Bulb}\left(t\right)-V_c\left(t\right)-\frac{Q\left(t\right)}{C}=0$

  $V_B-I\left(t\right)\times R-\frac{Q\left(t\right)}{C}=0$

Conclusion:

The voltage difference between the terminals of the battery is $V_{Battery}$ which is constant and the voltage difference across the capacitor $V_c\left(t\right)=\frac{Q\left(t\right)}{C}$ and across the bulb is $V_{Bulb}\left(t\right)=I\left(t\right)\times R$ .

To determine

Charge distribution on the plates of the capacitor.

Answer to Problem 1aT

One of the plates accumulates positive and other negative charge in equal magnitude.

Explanation of Solution

Introduction:

The charge on the capacitor is directly proportional to the potential difference across the capacitor plates,

  $\begin{array}{l}Q\propto V\\ Q=C\times V\end{array}$

Where ‘C’ is the constant known as the capacitance which depends on the material and design property of the capacitor.

In given circuit, shown in Figure 3, the positive terminal of the battery is connected with the upper plate of the capacitor and the lower one with negative terminal of the battery.

  

Figure 3: Charge distribution on the capacitor

Current flows from positive terminal of the battery to towards the bulb. Basically, current is in the opposite direction of the flow of the electrons. Electrons from the upper plate of the capacitor starts moving towards positive terminal of the battery and leaves the upper plate positive and electrons in the lower plate are repelled from the negative terminal of the battery. This accumulation of the charge happens till the potential difference across the plate is equal to the battery voltage. Also, the charge on the plates is equal in magnitude and opposite in charge.

Conclusion:

Hence, one of the plates of the capacitor accumulates positive and other negative charge in equal magnitude.

To determine

To Check: The predictions using voltmeter in the circuit.

Explanation of Solution

Introduction:

Ohm’s Law: The current in the circuit is directly proportional to the potential difference and the constant of proportionality is known as resistance R.

  $V=I\times R$

The charge on the capacitor is directly proportional to the potential difference across the capacitor plates,

  $\begin{array}{l}Q\propto V\\ Q=C\times V\end{array}$

Where ‘C’ is the constant known as the capacitance which depends on the material and design property of the capacitor.

In given circuit, shown in Figure 4, voltmeters are connected parallel to the battery, bulb and the capacitor in order to observe the potential for each circuit element after capacitor is fully charged. Once the capacitor is fully charged, circuit becomes open circuit and hence, no current flows through the circuit.

  

Figure 3: Circuit to calculate Vpotential across battery, bulb and the capacitor

Let say the battery has a $V_B=5\text{V}$ andthe resistance of the bulb is 5 ohmsand the capacitance of the capacitor is $5\text{μF}$ .

After a long time,the battery voltage is dropped cross the capacitor andpotential across the capacitor is calculated as 5 V and the voltmeter 3 reads 5 V.

Potential across the bulb, and the voltmeter 2 reads 0 V.

Voltmeter 1 reads 5 V.

Charge on the capacitor will be:

  $\begin{array}{c}Q=CV\\ =5\text{μF}\times \text{5V}\\ \text{=25}\mu \text{C}\end{array}$

Conclusion:

Hence, potential across the battery and the capacitor is 5 V and the bulb is 0 V. Charge on the capacitor is $25\mu C$ .