Question

ln the figure below you see a circuit with a battery (V), a capacitor (C), and two resistors (R1 and R2). The circuit has a switch that can be in two positions: switched to touch point A or switched to touch point B. By setting first the switch to A you can charge the capacitor, and then switching it to B you discharge the capacitor through the resistor. (This is the simple mechanism behind a camera flash, you can imagine the resistor R2 as a lightbulb that would "flash" for a short time when there is a current through it.)

Switching In the figure below you see a circuit with a battery (V), a capacitor (C), and two resistors (Rand R2). The circuit has a switch that can be in two positions: switched to touch point A or switched to touch point B. By setting first the switch to A you can charge the capacitor, and then switching it to B you discharge the capacitor through the resistor. (This is the simple mechanism behind a camera flash, you can imagine the resistor R2 as a lightbulb that would "flash" for a short time when there is a current through it.) R. - C - ER a) (1 point) Assume the capacitor is initially uncharged, you now close the switch to position A. What is the potential difference across the capacitor after a long time that the switch has been closed? b) (2 points) What is the charge on the capacitor after the switch has been closed to position A for a long time?

Answer 1

a potential difference across the capacitor after a long time. I after a long time there will be no current in circuit, so potential difference across capacitor will be [Ve=v 6) capacitor will be charged with charge a Q = CV c = cv c) switch to position B. for this V = VR₂ = I R2 2(t) = -da R at at 92 Ve RSVR

2=Q eft/RC) Ilt=/da/ Q & El RC Tour Tat - RC when switch has just been closed to

d) after a long time current through R2 It)= I e IRC to (I=0 e) I(t) = I et IRE 11. = VR2 It) f) 9 = Q et RC +