Question

A four-capacitor circuit is shown at right. Assume all four capacitors have different values of capacitance. 1. C2 and Cz are connected in series / parallel (circle one). Find their equivalent capacitance. v + c 2. Cz and C3 have the same / different (circle one) voltage drop across them. 3. C2 and C2 have the same / different (circle one) charge on their plates. A four-capacitor circuit is shown at right. Assume all four capacitors have different values of capacitance. 4. C2 and Cz are connected in series / parallel (circle one). Find their equivalent capacitance. 5. C2 and C3 have the same / different (circle one) voltage drop across them. 6. C2 and Cz have the same / different (circle one) charge on their plates.

Answer 1

For 1st Circuit

The capacitors C₁and C₄ are connected in series with voltage V. and the capacitors C₂ and C₃ are connected in parallel.

The combination of C₂ and C₃, i.e., C_2,3 (let) is connected in series with C₁ and C₄

(1)  The capacitors C₂ and C₃ are connected in parallel as the charge supplied by the voltage source gets divided into two parts, let's say 'Q₂' and 'Q₃' respectively.

And during parallel combination of the capacitors, the voltage dropped is same across each of them. Let the voltage drop across C₂ and C₃ be 'V₂' and 'V₃', respectively.

Now, the total charge across both the capacitors is given by

  

$\because Q = C \times V$

$\Rightarrow Q_{total} = C_{equivalent} \times V_{total} = (C_{2} \times V_{2}) + (C_{3} \times V_{3})$

$\because V_{total} = V_{2} = V_{3} = V$ ....................(for parallel combination)   

  $\Rightarrow Q_{total} = C_{equivalent} \times V = (C_{2} \times V) + (C_{3} \times V)$

  $\mathbf{\Rightarrow C_{equivalent} = C_{2,3} = C_{2} + C_{3} }$  

(2)  Capacitors have the same voltage across them, i.e.,

V₂ = V₃

(3) C₂ and C₃ have different charges on their plates.

FOR 2nd circuit

The capacitors C₂ and C₃ are connected in series and the combination of C₂ and C₃, i.e., C_2,3 (let) is connected in parallel with C_4.

The combination of C₂, C₃ and C₄ i.e., C_2,3,4 (let) is connected in series with C_1.

During series combination of the capacitors, charges stored in their plates are the same but the voltage drop across each of them is different.

Let the charges stored in the plates of capacitors C₂ and C₃ be 'Q₂' and 'Q₃' respectively and the voltage dropped across each of them is 'V₂' and 'V₃', respectively.

Therefore, Q₂ = Q₃

(a) The capacitors C₂ and C₃ are connected in series and the total voltage across both the capacitors is the sum of their individual voltage dropped and is given by

$\mathbf{V_{total} = V_{2}+V_{3}}$

$\because V = Q/C$

$\Rightarrow V_{total} =\frac{ Q}{C_{equivalent}} = \frac{Q_{2}}{C_{2}} +\frac{Q_{3}}{C_{3}}$

Since, Q₂ = Q₃ = Q

$\Rightarrow \frac{ Q}{C_{equivalent}} = \frac{Q}{C_{2}} +\frac{Q}{C_{3}}$

$\mathbf{\therefore \frac{ 1}{C_{equivalent}} =\frac{ 1}{C_{2,3}} = \frac{1}{C_{2}} +\frac{1}{C_{3}}}$