Question

Two capacitors of capacitances 1.0 microfarad and 2.0 microfarads are each charged by being connected across a 5.0-volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0 microfarad capacitor?​

Answer 1

Answer:

option c 1.7v

Explanation:

explanation in above picture please go through that

Answer 2

Given:

C1 = 1 μF

c2 = 2  μF

V = 5 V

To find:

Final voltage = V' =?

Explanation:

Two capacitors of 1 μF & 2 μF are first connected across a 5 V battery.

These capacitors are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together.

i.e. they are connected in a series combination.

The total capacitance in a series combination is given by

$\displaystyle \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$

     =   $\displaystyle \frac{1}{1\times 10^{-6}} + \frac{2}{1\times 10^{-6}}$

$\displaystyle \frac{1}{C_s} = \frac{3}{2}$

$\displaystyle \mathbf{ C_s = \frac{2}{3}}$

By comparing with the initial capacitance it is very easy to see that the series capacitance on 2μF when in series with 1μF.

∴ V' = $\frac{5}{3}$

V' = 1.666V

V'  = 1.7 V