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

Explanation:

The give data is as follows.

             C = 6 $\mu F$ = $6 \times 10^{-6} F$

             V = 6 V

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

              Q = CV

                  = $6 \times 10^{-6} F \times 6 V$

                  = $36 \times 10^{-6} C$

or,               = 36 $\mu C$

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

Answer 2

Answer:

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