In the circuit of Fig. 27-46,8=1.2 kV, C=6.5 μF, R1= R2=R3=0.73 M Ω. With C completely uncharged

step 1 In the circuit, EMF is 1 .2 kilowolts, capacitance 6 .5 micro ferrat. R1, R2 and R3 have the sam

In the circuit of Fig. 27-46,8=1.2 kV, C=6.5 μF, R1=...

In the circuit of Fig. $27-46,8=1.2 \mathrm{kV}, C=6.5 \mu \mathrm{F}, R_{1}=$ $R_{2}=R_{3}=0.73 \mathrm{M} \Omega$. With $C$ completely uncharged, switch $\mathrm{S}$ is suddenly closed (at $t=0$ ). At $t=0$, what are (a) current $i_{1}$ in resistor 1, (b) current $i_{2}$ in resistor 2, and $(\mathrm{c})$ current $i_{3}$ in resistor $3 ?$ At $t=\infty$ (that is, after many time constants), what are (d) $i_{1}$, (e) $i_{2}$, and (f) $i_{3} ?$ What is the potential difference $V_{2}$ across resistor 2 at $(g) t=0$ and (h) $t=\infty$ ? (i) Sketch $V_{2}$ versus $t$ between these two extreme times

Key Concepts

RC Circuit Transients

RC circuits combine resistors and capacitors, and their behavior in response to a sudden change in the circuit (such as closing a switch) is characterized by a transient response. This transient phase involves the capacitor charging or discharging over time, during which currents and voltages change according to the circuit’s time constant.

Time Constant

The time constant, typically denoted by ? (tau), is the product of resistance and capacitance in an RC circuit (? = RC). It determines the rate at which the capacitor charges or discharges, with the circuit reaching near steady-state after a few multiples of ?.

Initial Conditions in Capacitive Circuits

At the moment of a sudden change, such as a switch closure at t = 0, the capacitor’s initial condition plays a critical role. An initially uncharged capacitor behaves like a short circuit, meaning it momentarily allows maximum current flow, while its voltage is zero. This greatly affects the distribution of currents in the network immediately after the change.

Steady-State Behavior of Capacitor Circuits

In the steady state (as t approaches infinity), a capacitor in an RC charging circuit becomes fully charged and essentially behaves as an open circuit. At this point, no current flows through the capacitor, and the circuit analysis relies on static resistive networks, where voltages and currents are determined by the resistive paths alone.

Kirchhoff’s Circuit Laws

Kirchhoff’s voltage and current laws are fundamental in analyzing circuits. Applying these laws allows one to set up the equations needed to calculate unknown currents and voltages in the circuit during transient and steady-state conditions by accounting for the conservation of energy and charge.

Ohm’s Law

Ohm’s Law, which relates the voltage across a resistor to the current flowing through it and its resistance (V = IR), is essential for computing the current in each branch of the circuit. It serves as a key tool in both transient and steady-state analyses to evaluate how voltage divides among resistors.

Voltage Divider Rule

The voltage divider rule is used in resistor networks to determine the potential difference across a particular resistor. This concept is especially useful in steady-state analysis when the capacitor acts as an open circuit, allowing the resistors to be treated as part of a simple voltage division arrangement.

Transcript

00:01 In the circuit, emf is 1 .2 kilowolts, capacitance 6 .5 micro ferrat.

00:07 R1, r2 and r3 have the same values which is equal to 0 .73 mega -oom.

00:13 With c completely uncharged, switch as is suddenly closed at t equals to 0.

00:19 At t equals to 0, what are the currents i1 in the resistor r1, b, i2 in resistor r2, see i3 in resistor r3 and at t equals to infinite what are i 1 i 2 and i 3 what is the potential difference v2 across the resistor r2 at t equals to 0 and at t equals to infinite then sketch v2 versus t between these two extreme times.

01:02 When the switch is closed at t equals to zero we can replace the capacitor with a wire since at the beginning the capacitor is uncharged so the maximum current will flow in it which mean it act like a wire.

01:15 So the circuit can be reduced and shown and the current i1 will be equals to emf by r equivalent and r equivalent is here r1 plus r2 and r3 are in parallel which which can be written as r2 plus r3.

01:36 Now r equivalent will further be written as r1 into r2 plus r3 plus r3 plus r2 plus r3.

01:48 Substitute this into this current equation emf into r2 plus r3 divided by r1 into r2 plus r3 plus r2 r3.

02:03 Substituting the values given in the question, 1 ,200 multiplied 2 into 7 .3 into 10 raised to the power 5 om divided by 3 into 0 .7 .3 into 10 raised to the power 5 ome square, where r equals 3 and the value from this we will get 0 .0011 ampere or 1 .1 millie ampere.

02:44 This is the value of i1...

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