Question

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 1

Concepts and reason

The concepts used to solve this problem are the series combination of resister’s concept, the parallel combination of the capacitor’s concept, the ohm’s law, and the time varying voltage concept.

Initially calculate the equivalent resistance of the two resistors by using the series combination of resister’s concept, and then calculate the equivalent capacitance of the two capacitors by using the parallel combination of the capacitor’s concept, and then calculate the time interval during which the voltage to reduce to its initial value by using the time varying voltage concept and then calculate the current through the circuit by using the time varying voltage concept.

Fundamentals

Two or more resistors are connected, and if the equivalent resistance of those resistors is the algebraic sum of the resistances of all the resistors, then the combination of the resistors , , … is called the series combination of all n resistors.

If the resistors , , … are n number of resistors and connected in series, the equivalent resistance of the resistors , , … is,

Here, is the equivalent resistance of all the n resistors.

Two or more capacitors are connected, and if the equivalent capacitance of those capacitors is the algebraic sum of the capacitances of all capacitors, then the combination of the capacitors, , , , …. is called the parallel combination of all n capacitors.

If the capacitors , , , …. are n numbers of capacitors and connected in parallel, then the equivalent capacitance of the n capacitors is,

Here, is the equivalent capacitance of all the n capacitors.

A circuit contains a capacitor of capacitance C and a resistor of resistance R and a switch S.

Before the switch S is closed, the voltage across the circuit is

After the switch is closed the time varying voltage is,

$V=Ve-RC $

Here, V is the voltage after the switch is closed, is the initial voltage, and t is the time interval.

The ohm’s law is,

Here, V is the voltage across the circuit before the switch is closed, I is the current through the circuit, and R is the resistance of the resistor.

Substitute for V in the equation and solve for I.

(a)

The resistors , and the capacitors , are arranged in the circuit is as shown in the figure.

From the circuit diagram, the two capacitors are in parallel combination then the equivalent capacitance of these capacitors is,

$C=C+C2 $

Here, is the equivalent capacitance of the two capacitors, is the capacitance of the left side capacitor, and is the capacitance of the right-side capacitor

Substitute for and for in the equation $C=C+C2 $ and solve for .

The two resistors are in series combination, and then the equivalent resistance of these resistors is,

Here, is the equivalent resistance of the two resistors, is the resistance of the bottom resistor, and is the resistance of the top-right resistor.

Substitute for and for in the equation and solve for .

Before the switch S is closed, the voltage across the circuit is 45.0 V.

After the switch S is closed, the voltage changes to 10.0 V.

The voltage across the circuit, after the switch S is closed

$V=VeRS $

Here, V is the voltage across the circuit after the switch is closed, is the initial voltage, t is the time interval during which the voltage to reduce to its initial value.

Rewrite the equation for t.

Substitute for , for , for V, and 45.0 V for in the above equation and solve for t.

(b)

The current through the circuit, after the switch S is closed,

Here, I is the current through the circuit after the switch is closed.

Substitute 45.0 V for , for , for , and for t in the equation and solve for I.

Ans: Part a

Thus, the time interval during which the voltage to reduce to its initial value is .

Part b

Thus, the current through the circuit at the time t is 0.125 A.