Question

In the figure given below, the resultant capacitance between points \(A\) and \(B\) is:

• A. \(20\,\mu F\)
• B. \(\dfrac{20}{3}\,\mu F\)
• C. \(\dfrac{40}{3}\,\mu F\)
• D. \(10\,\mu F\)

Hint:

For a balanced capacitor bridge: \[ \frac{C_1}{C_2} = \frac{C_3}{C_4} \] the bridge capacitor carries no charge and can be removed from the circuit. This greatly simplifies capacitor-network problems.

Answer 1

The Correct Option is A

Label the junction between \(10\,\mu F\) and \(5\,\mu F\) as \(X\), and the junction between \(5\,\mu F\) and \(40\,\mu F\) as \(Y\). Step 1: {Identify the bridge balance condition.} The capacitor network has: \[ \frac{10}{20} = \frac{20}{40} = \frac{1}{2} \] Since the ratios are equal, the bridge is balanced. Hence no potential difference exists across the central capacitor. Therefore the \[ 5\,\mu F \] capacitor can be ignored. Step 2: {Reduce the upper branch.} Upper branch consists of: \[ 20\,\mu F \] in series with \[ 40\,\mu F \] Thus, \[ C_{\text{upper}} = \frac{20\times40}{20+40} \] \[ = \frac{800}{60} \] \[ = \frac{40}{3}\,\mu F \] Step 3: {Reduce the lower branch.} Lower branch consists of: \[ 10\,\mu F \] in series with \[ 20\,\mu F \] Therefore, \[ C_{\text{lower}} = \frac{10\times20}{10+20} \] \[ = \frac{200}{30} \] \[ = \frac{20}{3}\,\mu F \] Step 4: {Combine the two branches.} The upper and lower branches are in parallel. Hence, \[ C_{\text{eq}} = C_{\text{upper}} + C_{\text{lower}} \] \[ = \frac{40}{3} + \frac{20}{3} \] \[ = \frac{60}{3} \] \[ = 20\,\mu F \] Therefore, the resultant capacitance between \(A\) and \(B\) is \[ \boxed{20\,\mu F} \]