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
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
b) What will be the current at that time?
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 nresistors.
have n resistors connected in series, then the equivalent resistance of
represents the equivalent resistance to all nresistors.
If two or more capacitors have been connected, the equivalent capacitance is
refers to the parallel combination all ncapacitors.
If there are nnumbers (or more) of capacitors connected in parallel, the equivalent capacitance for the ncapacitors will be,
This is the equivalent capacitance for all ncapacitors.
A circuit includes a capacitor of capacitance and a resistor R, as well as a switch S.
The time varying voltage will be,
represents the initial voltage and T the time interval.
The ohm’s law is
Vis voltage across circuit before switch is closed. Iis current through circuit. Ris resistance of resistor.
Substitute HTMLmedia_tag_24$ in the equation to solve for I.
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.
represents the capacitance for the left-side capacitor and #media_tag_30$ the capacitance for the right-side capacitor.
within the equation
, and then solve for
These resistors can be used in series, so the equivalent resistance is.
refers to the resistance of bottom resistor and #media_tag_46$ refers to the resistance for top-right resistor.
, 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
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
respectively for V.
After the switch S has been closed, the current flows through the circuit.
Iis here the current flowing through the circuit after it is closed.
Substitute 45.0V for
in the equation