Question

Three 1.5 V cells are arranged in series, powering a circuit that has two operations with equivalent resistances of 104 Ω (connecting to the positive terminal) and 411 Ω that meet in a single, metered connection. There is a 22 μC capacitor that bridges between the connection and the negative terminal.What is the potential at the connection point(in Volts) What is the charge on the capacitor?

Answer 1

effective voltage = 1.5+ 1.5+1.5 = 4.5 V

For a D.C. voltage capactor behaves like an open circuit.

so,

Potential at the connection will be same as the potential of cell

which is equal to 4.5 V

so

potential at connection = $4.5\ V$

Charge on capacitor = $Q=CV$

$C=22\ \mu C\\V=4.5\ V\ so\\ Q=22 \times 4.5=99\ \mu\ C$