Question

My Notes (a) A 41 Ω resistor is connected in series with a 6 µF capacitor and a battery. What is the maximum charge to which this capacitor can be charged when the battery voltage is 6 V? (When entering units, use micro for the metric system prefix µ.)

Answer 1

Step-by-step explanation:

capacitance, C = 6 μF

Voltage, V = 6 V

Let the maximum charge is Q.

Q = C x V

Q = 6 x 6 = 36 μC

Answer 2

Step-by-step explanation:

The give data is as follows.

C = 6
`\mu F` =
`6 * 10^(-6) F`

V = 6 V

Now, we know that the relation between charge, voltage and capacitor for series combination is as follows.

Q = CV

=
`6 * 10^(-6) F * 6 V`

=
`36 * 10^(-6) C`

or, = 36
`\mu C`

Thus, we can conclude that maximum charge of the given capacitor is 36
`\mu C`.