Right after S is moved to the position a, following a long time in the position

Question

Right after S is moved to the position a, following a long time in the position b, the current through R is /R.

If S is in the position a for a long time, the charge on C approaches zero.

If S is in the position a for a long time, the current through R approaches /R.

Right after S is moved to the position b, following a long time in the position a, the potential across C is nearly zero.

Right after S is moved to the position b, following a long time in the position a, the current through R is zero.

After S is moved to b, it takes a time = RC, for the charge on C to drop to 1/e of its initial value.

Right after S is moved to the position b, following a long time in the position a, the current through R is /R.

Explanation / Answer

Right after S is moved to the position a, following a long time in the position b, the current through R is /R.

True : as the Switch is moved to "a" , the capacitor behaves as a short circuit since it has no charge or voltage across it. hence all the voltage appears across the resistor and using ohm;s law , current is given as

i = E/R

After S is moved to b, it takes a time = RC, for the charge on C to drop to 1/e of its initial value.

True : since Q =Qo e-t/RC

at t = RC ,

Q =Qo e-1 = Qo /e

Right after S is moved to the position b, following a long time in the position a, the current through R is /R

True :

after spending long time in "a" , the capacitor gets fully charged and Voltage across it becomes equal to battery Voltage "E". as the switch is moved to "b" , the resistor becomes in parallel with the capacitor and Voltage across capacitor = Voltage across resistor = E

hence using ohm;'s law , i = E/R

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.