Question 454: Electrostats & Capacitance

Question

Two capacitors of $100 \mu \mathrm{~F}$ and $50 \mu \mathrm{~F}$ are connected in parallel. If the potential difference across $100 \mu \mathrm{~F}$ is 20 V and across $50 \mu \mathrm{~F}$ is 40 V , then the common potential of the parallel combination will be (same polarities of the capacitor connected together)

Accepted Answer

Step 1: Find the total charge before connecting.
Total charge before connecting = charge on $C_1$ + charge on $C_2$:

$Q_	ext{total} = C_1 V_1 + C_2 V_2$
Step 2: After connecting, both capacitors reach a common voltage $V$.
Total charge = $(C_1 + C_2) V$
Step 3: Set the total charges equal (since charge is conserved):
$$(C_1 + C_2) V = C_1 V_1 + C_2 V_2$$

So,

$$V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$
Step 4: Plug in the values.
$C_1 = 100 	imes 10^{-6}\ \mathrm{F}$, $V_1 = 20\ \mathrm{V}$

$C_2 = 50 	imes 10^{-6}\ \mathrm{F}$, $V_2 = 40\ \mathrm{V}$
$V = \frac{100 	imes 10^{-6} 	imes 20 + 50 	imes 10^{-6} 	imes 40}{100 	imes 10^{-6} + 50 	imes 10^{-6}}$
Step 5: Calculate:
$100 	imes 10^{-6} 	imes 20 = 2 	imes 10^{-3}$

$50 	imes 10^{-6} 	imes 40 = 2 	imes 10^{-3}$

Total numerator $= 4 	imes 10^{-3}$
$100 	imes 10^{-6} + 50 	imes 10^{-6} = 150 	imes 10^{-6} = 1.5 	imes 10^{-4}$
$V = \frac{4 	imes 10^{-3}}{1.5 	imes 10^{-4}} = \frac{400}{15} = \frac{80}{3}$

Answer

20 V

Answer

60 V

Answer

$\frac{3}{80}$

Answer

$\frac{80}{3} \mathrm{~V}$

Electrostats & Capacitance

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