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QPages 165-170: Q14: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q14 Solution: It is required to construct a 10 �F capacitor which can be…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often think that putting capacitors in series preserves the 10 µF value while simply adding the voltage ratings. In reality, series increases the voltage rating but the equivalent capacitance becomes $C_{ ext{eq}}=10/4=2.5,mu ext{F}$. To recover the desired 10 µF, you must add enough such series groups in parallel—here, 4 rows of 4 capacitors each—so that the total capacitance is $4 imes2.5=10,mu ext{F}$ while the voltage rating reaches $4 imes50=200$ V.

It is required to construct a 10 �F capacitor which can be connected across a 200 V battery. Capacitors of capacitance 10 �F are available but they can withstand only 50 V. Design a combination which can yield the desired result.

Step-by-Step Verified Solution

**Answer:** 4 rows, each with 4 capacitors of 10 _F. **Step-by-Step Explanation:** To withstand 200 V with capacitors rated at 50 V, we need xx in series such that x×50≥200⇒x=4x \times 50 \geq 200 \Rightarrow x = 4. Equivalent capacitance of 4 in series: Crow=10/4=2.5C_{row} = 10/4 = 2.5 _F. To get 10 _F total, number of rows y=10/2.5=4y = 10/2.5 = 4. So, 4 rows, each with 4 capacitors .

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