**7) In the circuit shown in the figure both capacitors are initially charged to 45.0 V….**

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## Question “7) In the circuit shown in the figure both capacitors are initially charged to 45.0 V….”

7) In the circuit shown in the

figure both

capacitors are initially charged to 45.0 V.

a) How long after closing the switch S will the

potential across each capacitor be reduced to

10.0 V?

b) What will be the current at that time?

## Answer

This problem can be solved using the following concepts: the series combination resister’s idea, the parallel combination capacitor’s notion, the ohms law and the time varyingvoltage concept.

First, calculate the equivalent resistance between the resistors using the series combination resistor’s concept. Next, calculate the equivalent capacitance between the two capacitors using the parallel combination capacitor’s idea. Then calculate the time period during which the voltage will drop to its initial value using the time-varying voltage concept. Finally, calculate the current through a circuit using the time-varying voltage concept.

If two or more resistors have been connected, the equivalent resistance of each resistor is the algebraic sum all resistors. The combination of resistors and is known as the series combination all *n*resistors.

If

and

have n resistors connected in series, then the equivalent resistance of

and

respectively is

.

represents the equivalent resistance to all *n*resistors.

If two or more capacitors have been connected, the equivalent capacitance is

, HTMLmedia_tag_12$ and HTMLmedia_tag_13$. The combination of these capacitors, , …. is called the parallel combination of all i>n/i>capacitors.

refers to the parallel combination all *n*capacitors.

, HTMLmedia_tag_12$ and HTMLmedia_tag_13$ are connected in parallel, then the …. capacitors .

If there are *n*numbers (or more) of capacitors connected in parallel, the equivalent capacitance for the *n*capacitors will be,

This is the equivalent capacitance for all *n*capacitors.

A circuit includes a capacitor of capacitance and a resistor *R*, as well as a switch S.

The voltage across the circuit before the switch S is shut is

The time varying voltage will be,

*V* indicates the voltage after the switch has been closed.

represents the initial voltage and *T* the time interval.

The ohmâ€™s law is

*V*is voltage across circuit before switch is closed. *I*is current through circuit. *R*is resistance of resistor.

Substitute HTMLmedia_tag_24$ in the equation to solve for *I*.

(a)

The circuit layout is shown in Figure.

The circuit diagram shows that the two capacitors are connected in parallel. This is because the equivalent capacitance of these capacitors can be seen.

refers to the equivalent capacitance between the two capacitors.

represents the capacitance for the left-side capacitor and #media_tag_30$ the capacitance for the right-side capacitor.

Substitute

to

,

to

within the equation

, and then solve for

.

These resistors can be used in series, so the equivalent resistance is.

represents the equivalent resistance to the two resistors.

refers to the resistance of bottom resistor and #media_tag_46$ refers to the resistance for top-right resistor.

Substitute

to

,

to

in equation

, and then solve for

.

The circuit voltage is 45.0V before the switch S is shut.

The voltage drops to 10.0V after the switch S has been closed.

After the switch S has been closed, the voltage across the circuit is zero

*V* is the voltage across the circuit when the switch is closed.

indicates the initial voltage. *T* indicates the time period during which the voltage will decrease to its initial value.

Rewrite the equation *to*.

In the equation above, substitute

to

;

to

;

and

respectively for V.

(b)

After the switch S has been closed, the current flows through the circuit.

*I*is here the current flowing through the circuit after it is closed.

Substitute 45.0V for

;

to

;

to

;

to

in the equation

.