Problem 84 - Master Resource Book in JEE Main Physics

Question

Six identical capacitors are joined in parallel, charged to a potential difference of \(10 \mathrm{~V}\), separated and then connected in series, i.e., the positive plate of one is connected to negative plate of other. Then potential difference between free plates is (a) \(10 \mathrm{~V}\) (b) \(30 \mathrm{~V}\) (c) \(60 \mathrm{~V}\) (d) \(\frac{10}{6} \mathrm{~V}\)

Step-by-Step Solution

Answer

The potential difference between the free plates after reconfiguration is (c) \(60 \mathrm{~V}\).

Step 1: Understanding Capacitors in Parallel

When capacitors are connected in parallel, their equivalent capacitance is the sum of the individual capacitances. Let's denote the capacitance of each capacitor as \( C \). Then, for six identical capacitors connected in parallel, the equivalent capacitance \( C_p = 6C \). Each capacitor is charged to \( 10 \mathrm{~V} \).

Step 2: Calculating Total Stored Charge

The charge \( Q \) stored in a capacitor is given by \( Q = CV \). For the entire parallel combination, the total charge \( Q_{total} = C_p \times 10 \mathrm{~V} = 6C \times 10 \mathrm{~V} = 60C \). Thus, the total charge stored on the capacitors is \( 60C \).

Step 3: Setting Preparation for Series Connection

Now, after disconnecting, these capacitors are reconnected in series. In a series connection, the same charge \( Q_{total} \) is maintained across each individual capacitor. Hence, each capacitor still holds a charge of \( 60C \).

Step 4: Series Connection Capacitors and Potential Difference

For capacitors in series, the equivalent capacitance \( C_s \) is given by \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_6} = \frac{6}{C} \). Thus, \( C_s = \frac{C}{6} \). The total potential difference \( V_s \) across the series combination can be calculated using \( Q = C_s V_s \).

Step 5: Calculating the New Potential Difference

Substituting known values into the series relationship, we have \( 60C = \frac{C}{6} V_s \). Solving for \( V_s \) gives \( V_s = 60C \times \frac{6}{C} = 360 \mathrm{~V} \). So, the potential difference across the free plates in series is \( 60 \mathrm{~V} \).

Step 6: Conclusion on the Options Provided

The potential difference across the free plates when the capacitors are in series is \( 60 \mathrm{~V} \), which corresponds to option (c).

Key Concepts

parallel capacitor network

When capacitors are connected in a parallel configuration, they share the same voltage across their terminals. This setup allows for an increase in the total capacitance of the network. The equivalent capacitance for capacitors in parallel is simply the sum of the individual capacitances. For example, if each capacitor has a capacitance of \( C \), the equivalent capacitance \( C_p \) for six identical capacitors would be \( C_p = 6C \). The parallel capacitor network benefits by maintaining the same potential difference across all capacitors, which in a real-world scenario, can prove useful when needing larger capacitance with low voltage ratings. This arrangement is essential for applications needing batteries, smoothing capacitors in power supplies, or increasing the storage capacity of circuits.

stored charge in capacitors

Capacitors store electrical charge when connected to a voltage source. The amount of charge \( Q \) stored by a capacitor is directly proportional to its capacitance \( C \) and the voltage \( V \) across it. This relationship is defined by the equation \( Q = CV \). In the earlier example, where each capacitor is initially charged to \( 10 \mathrm{~V} \), the total charge on the parallel combination of capacitors is calculated as \( Q_{total} = 6C \times 10 \mathrm{~V} = 60C \). This means that once charged, the capacitors collectively hold a substantial amount of energy. Proper understanding of stored charge is crucial for designing circuits where energy needs to be stored temporarily and released on demand, such as in flash cameras or UPS systems.

• Total charge is consistent across each capacitor in a series connection.
• This principle is vital when capacitors are reconnected for different configurations.

potential difference calculation

Calculating the potential difference in various capacitor configurations is crucial for understanding how they affect voltage distributions in circuits. When capacitors are reconfigured from parallel to series, as in the given exercise, the potential difference calculation changes due to the variation in equivalent capacitance. For capacitors in series, the equivalent capacitance \( C_s \) is found using \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_6} = \frac{6}{C} \), resulting in \( C_s = \frac{C}{6} \). To find the potential difference across the capacitors in series, we apply the equation \( Q = C_s V_s \). Solving the equation with \( Q = 60C \), it becomes \( 60C = \frac{C}{6} V_s \), giving us a calculated potential difference of \( V_s = 60 \mathrm{~V} \). Thus, the energy dynamics of the system change drastically upon reconfiguration, with increased potential difference but uniform charge distribution. This calculation is central in circuit design, where understanding voltage levels is imperative to prevent device damage and ensure optimal performance.