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:

1.67 V

Explanation:

Q = Capacitance

Q' = Capacitance after connecting in series

V = Voltage

$V_t$ = Voltage of battery = 5 V

$Q_1=C_1\times V_t\\\Rightarrow Q_1= 1\times 5=5\ \mu F$

$Q_2=C_2\times V_t\\\Rightarrow Q_2= 2\times 5=10\ \mu F$

As the plates opposite charge are connected together they are in series

$Q_1'=Q_2'=Q_t$

$\frac{1}{C_t}=\frac{1}{C_1}+\frac{1}{C_2}\\\Rightarrow \frac{1}{C_t}=\frac{1}{1}+\frac{1}{2}\\\Rightarrow \frac{1}{C_t}=\frac{3}{2}\\\Rightarrow C_t=\frac{2}{3}$

$Q_2'=C_tV_t\\\Rightarrow Q_2'=\frac{2}{3}\times 5\\\Rightarrow Q_2'=\frac{10}{3}\ \mu F$

$V_2=\frac{Q_2'}{C_2}\\\Rightarrow V_2=\frac{\frac{10}{3}}{2}\\\Rightarrow V_2=1.66\ V$

The voltage across the 2 microfarad capacitor is 1.67 V