In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 uF. All

Q: Why is capacitor C in series with the rest of the network?

Because all the charge flowing from terminal A must pass through capacitor C before it splits into the various branches of the network.

Q: How do we know which capacitors are in parallel or series in the complex network?

Look for nodes where the current splits (parallel) or where components are connected end-to-end with no other connections (series). Redrawing the circuit can often make these relationships clearer.

Q: Why did we split the network into left and right branches?

Complex bridge-like structures can sometimes be simplified by identifying parallel sub-networks. The solution uses this approach to break the problem into manageable pieces.

Q: What if the network was a balanced Wheatstone bridge?

If it were balanced, the capacitor in the middle branch would have no potential difference across it and could be removed, drastically simplifying the circuit. Always check for balance first!

Q: Why is the formula for series capacitors different from series resistors?

Capacitance is proportional to plate area and inversely proportional to distance. Connecting capacitors in series effectively increases the distance between the outermost plates, decreasing the overall capacitance.

Question

In the circuit shown, find $C$ if the effective capacitance of the whole circuit is to be $0.5\mu	ext{F}$. All values in the circuit are in $\mu	ext{F}$.

Options:

(A) $\frac{6}{5}\mu	ext{F}$

(B) $4\mu	ext{F}$

(C) $\frac{7}{10}\mu	ext{F}$

(D) $\frac{7}{11}\mu	ext{F}$

Sigma Percentile · Accepted Answer

### Demystifying Complex Capacitor Networks

Welcome back, circuit solvers! Today we are tackling a classic capacitor network problem that looks like a tangled web at first glance. But don't let the complexity intimidate you. The secret to solving these kinds of problems is the 'divide and conquer' strategy. We will break this intimidating circuit down into smaller, manageable chunks.

### Breaking it Down: The Divide and Conquer Strategy

Look closely at the circuit diagram. The very first thing you should notice is the position of the unknown capacitor, **$C$**. It sits right at the entrance of the circuit, connected in series with the entire complex network to its right.

This is a crucial observation. It means our primary mission is to find the equivalent capacitance of that entire messy network. Once we have that single equivalent value, finding $C$ becomes a straightforward series calculation.

### Simplifying the Network Branches

The network itself can be conceptually split into two main parallel branches connected between the primary upper and middle nodes.

Let's focus on the **right section** first. By carefully tracing the paths and applying standard series and parallel simplification rules (remember, capacitors in series add like resistors in parallel, and vice versa!), this right branch elegantly reduces to an equivalent capacitance of **$1\mu	ext{F}$**.

Next, we shift our attention to the **left section**, which encompasses the remaining capacitors. Applying the same systematic simplification techniques to this sub-circuit yields an equivalent capacitance of **$\frac{4}{3}\mu	ext{F}$**.

### Bringing it Together

Now the magic happens. Since these two simplified branches are connected in parallel, we simply add their capacitances together to find the total equivalent capacitance of the network ($C_{net}$):

$$C_{net} = C_{left} + C_{right}$$
$$C_{net} = \frac{4}{3} + 1 = \frac{7}{3}\mu	ext{F}$$

### The Final Equation

We are almost there! We now know that the entire complex network behaves exactly like a single $\frac{7}{3}\mu	ext{F}$ capacitor. And remember our first observation? This network is in series with our unknown capacitor $C$.

The problem states that the total effective capacitance of the whole circuit is $0.5\mu	ext{F}$ (or $\frac{1}{2}\mu	ext{F}$). We can now set up our final series capacitance equation:

$$C_{eq} = \frac{C \cdot C_{net}}{C + C_{net}} = \frac{1}{2}$$

Substituting our value for $C_{net}$:

$$\frac{C \cdot (7/3)}{C + 7/3} = \frac{1}{2}$$

Now, it's just a matter of simple algebra. Cross-multiply to solve for $C$:

$$\frac{14}{3}C = C + \frac{7}{3}$$
$$\frac{14}{3}C - \frac{3}{3}C = \frac{7}{3}$$
$$\frac{11}{3}C = \frac{7}{3}$$

And finally, we arrive at our answer:

**$$C = \frac{7}{11}\mu	ext{F}$$**

By breaking the problem down step-by-step, what looked like a nightmare became a highly logical and satisfying puzzle to solve!

Animated Solution for Physics - Electrostatics: In the circuit shown, find $C$ if the effective capacitance of the whole circuit is to be $0.5 μ F$ . All values in the circuit are in $μ F$ .

Select Answer:

Visualized Solution

$C_{e q}$ of the Network

Simplifying the Right Branch

Simplifying the Left Branch

Total Network Capacitance

Setting up the Final Equation

Solving for C

The Sigma Insight: Combination of Capacitors

Demystifying Complex Capacitor Networks

Breaking it Down

Simplifying the Network Branches

Bringing it Together

The Final Equation

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