Question

What is the Kirchhoff's junction rule equation for this junction? A 10-V-emf battery is connected in series with the following: a 2-mu F capacitor, a 2-Ohm resistor, an ammeter, and a switch, initially open; a voltmeter is connected in parallel across the capacitor. After the switch has been closed for a relatively long period (several seconds, say), what are the current and capacitor voltage readings, respectively? A resistor, inductor, battery, and switch are all hooked up in series. Draw a graph showing the current through the resistor as a function of time, with t = 0 representing the moment just after the switch is closed. Be sure to label the axes of the graph. If the circuit above (1c) has resistance 10 Ohms, inductance 50 milliHenry, and battery of 24 Volts, what is the maximum current in the circuit?

Answer 1

Q1.a )

kirchoff's current law states that:

total current entering at a node=total current leaving a node

here total current entering=0 A

total current leaving =I1+I2+I3

hence KCL states that:

I1+I2+I3=0

hence option d is correct.

Q1.b)

time constant of the circuit=R*C=2*2*10^(-6)=4 us

hence several seconds is enough for the circuit to reach steady state.

at steady state, capacitor acts like an open circuit

hence after several seconds of switch being closed, capacitor will be open circuit

and hence current in the circuit=0

capacitor will be charged to the connected potential=10 volts

hence ammeter reading is 0 A and voltmeter reading is 10 volts.

hence option a is correct.

Q1.c)

as inductor behaves like an open circuit when the switch is just closed,

initial current=0 A

after steady state is reached, inductor behaves like a short circuit

hence current in the circuit=V/R

the current will rise exponentially

hence current at any time t is given as

i(t)=(V/R)*(1-exp(-t/T))

where T=L/R=time constant

graph:

current vs time 2.5 1.5 05- 0.005 0.01 0.015 0.02 0.04 0.025 time (in seconds 0.03 035 0.045 0.05

Q1.d)maximum current=steady state current=V/R=24/10=2.4 A