Question

In the diagram below,(Figure 1) the two resistors, R1 and R2, are identical and the capacitor is initially uncharged with the switch open.

Part A

How does the current through R_m1 compare with the current through R_m2 immediately after the switch is first closed?

The current through R_m1 Please Chooseis greater thanis less thanis equal tothe current through R_m2 .

Part B

How does the current through R_m1 compare with the current through R_m2 a very long time after the switch has been closed?

The current through R_m1 Please Chooseis greater thanis less thanis equal tothe current through R_m2 .

Part C

How does the current through R_m1 compare with the current through R_m2 immediately after the switch is opened(after being closed a very long time)?

The current through R_m1 Please Chooseis greater thanis less thanis equal tothe current through R_m2 .

Answer 1

(A) Kirchhoff's Node Rule is that, the sum of currents flowing towards any node in an electrical circuit is equal to the sum of currents flowing away from that node. From the above node rule, it follows that immediately after the switch is closed, the current Through resistor \(R_{1}\) is equal to the sum of currents through the resistor \(R_{2}\) and capacitor \(C\). Thus we have \(I_{R_{1}}=I_{R_{2}}+I_{C}\)

Hence, \(I_{R_{1}}>I_{R_{2}}\)

(B) When the switch is closed the capacitor gets charged and after the switch is closed for long time, the capacitor gets fully charged and hence no current flows through the capacitor. Thus we have \(I_{R_{1}}=I_{R_{2}}+I_{C}\)

\(=I_{R_{2}}+0\)

\(=I_{R_{2}}\)

Hence, \(I_{R_{1}}=I_{R_{2}}\)

(C)

Now after the switch is open, then the resistor \(R_{1}\) and the battery will be out of circuit. After the switch is opened immediately, the capacitor discharges through the resistor \(R_{2}\). Which means that Current flows only through the resistor \(R_{2}\).

Hence,

\(I_{R_{1}}

Answer 2

Concepts and reason

The concepts required to solve the problem is Kirchhoff’s law and current through capacitors.

Use Kirchhoff’s current law in a junction to compare the current immediately after the switch is closed. Use the charge flow through the capacitor to calculate the current through the capacitor when the switch is closed for a long time. Use the discharging of the capacitor to compare the current through the resistors when the switch is opened after being closed for a long time.

Fundamentals

According to Kirchhoff’s current law, the current entering into a junction of circuit is equal to the current leaving the junction.

The capacitor is a device used to store charge. While a capacitor is connected to a source voltage, the capacitor charges to the maximum and the flow of current, stops. When the charged capacitor is disconnected from the source voltage, the capacitor starts to discharge and the current flows through the circuit.

Immediately, after closing the switch, the resistors and capacitor are connected to the voltage source and current starts to flow through the circuit to charge the capacitor.

According to Kirchhoff’s current law or Kirchhoff’s junction rule, the current entering into a junction of circuit is equal to the current leaving the junction of circuit.

Consider the junction of circuit which connecting the two resistors and the capacitor. The Kirchhoff’s current equation for the junction is,

${I_{{R_1}}} = {I_{{R_2}}} + {I_C}$

Here, ${I_{{R_1}}}$ is the current through the first resistance that enters into the junction, ${I_{{R_2}}}$ is the current through the second resistor that leaves the junction and ${I_C}$ is the current through the capacitor that leaves the junction. Thus,

${I_{{R_1}}} > {I_{{R_2}}}$

The current through the first resistor is greater than the current through the second resistor immediately after closing the switch.

According to Kirchhoff’s current rule, the current at the junction when the resistors and capacitors are connected to the source voltage is,

${I_{{R_1}}} = {I_{{R_2}}} + {I_C}$

While the switch is closed for a long time, the capacitor charges to the maximum. After the capacitor gets charged to maximum, there will not be current flow in the capacitor. The current through capacitor is,

${I_C} = 0$

Thus the current through the resistors is,

${I_{{R_1}}} = {I_{{R_2}}}$

When the switch is closed for a long time, the current through the resistors are same.

When the switch is open, the capacitor starts discharging. The current flows through the loop which include the second resistor. Then the current through the second resistor will be greater than the current through the first resistor.

${I_{{R_2}}} > {I_{{R_1}}}$

When the switch is open, the current through the second resistor will be greater than the current through the first resistance.

Ans:

The current through the first resistor is greater than the current through the second resistor immediately after closing the switch.

When the switch is closed for a long time, the current through the resistors are same.

When the switch is open, the current through the second resistor will be greater than the current through the first resistance.