Question: A circult is represented below. Initially, the switch next to the capa

Question

Question: A circult is represented below. Initially, the switch next to the capacitor C is open, and the capacitor is uncharsed The electromotive force of the battery is known, as well as the values of the resistances, and of the capacitance C 1) Find the value of the currents 1.1. 12. I.1? when the switch is open. 2) Find the potential difference among the points a and b represented in the figure. The switch now is closed, at ta0, and the switch is kept close for a long time Find: 3) The value of the currents 1. s. 1. I le right after the switch is closed. 4) The final yalue (le. the value after a long time has passed) of the current through each resistor,and of the charge on the capacitor NOTE: Expla 5) [OPTIONALJ find the dependence I(t), and Q(t) for each time t. 1(t) is the current provided by the battery, and Qt) lis the charge on the capacitor. Switch open 11 # Switch closed, t-0 Switch closed, after a long time has passed 4R 2)

Explanation / Answer

When the switch is open:

(1) Resistors 4R and 2R will be in series with equivalent resistance R1 = 6R and Resistors R and 5R will be in series, with equivalent resistance R2 = 6R.

Current I1 = I3 = E/6R

Similarly, I2 = I4 = E/6R

Since R1 and R2 are in parallel, their equivalent resistance is 6R.6R/12R=3R

Current I = E/3R

I1 = I3 = I2 = I4 = I/2 = E/6R

(2) Va - 2R.I3 - 5R.I4 - Vb =0

Vab = Va - Vb = 7R.E/6R = 7E/R

When the swit h is just closed, the capacitor will just behave as an short circuited element. So, points a and b will be at same potential. As a result, the resistances 4R and R will be in parallel with equivalent resistance R1 = 4R/5 and resistances 2R and 5R are in parallel, with equivalent resistance R2 = 10R/7. R1 and R2 are in series. So,

R' = 78R/35

I = E/R' = 35E/78R

Voltage across combination R1, V1 = I.R1 = 14E/39

Voltage across combination R2, V2 = I.R2 = 25E/39

I1 = V1/4R = 7E/78R

I2 = V1/R = 14E/39R

I3 = V2/2R = 25E/78R

I4 = V2/5R = 5E/39R

When the switch is closed for a long time

(4) A voltage equal to E will now appear across the capacitor

Resistances 4R and R will be in series now and Resistances 2R and 5R will be series.

Current, I1 = I2 = E/5R

Current I3 = I4 = E/7R

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