A 1.4 uF capacitor is connected in series to a 0.7 uF capacitor. The two are fur

Question

A 1.4 uF capacitor is connected in series to a 0.7 uF capacitor. The two are further connected to a 41.4 Volts non-ideal battery through a 1.6 M resistor. All connections are in series. The internal resistance of the battery is 9.8 m. Both capacitors are initially discharged.

The switch is closed at t=0 s. Find the current in the circuit at t=5.0 seconds.

Note: the current in the circuit after a few seconds is likely to be very small. It is likely that just putting the numbers into a calculator may give you zero, which will not be the acceptable answer. Calculate the result in nA instead.

Explanation / Answer

the growth decay of charge in the circuit given by qgrowth=q0(1-e-t/T) and q=qo(e-t/T)   

T=time constant=RC (1.6(106)+9.8(10)-3=1.60000000098s )(.7)(.14)/.7+.14 (10-6)= 0.0752(10)-6  

this situation concerns with growth of charge and constitues some current i=i0(-e-t/RC)

i0=E0/R=41'4/1.60000000098=41.4/1.6=.3906A=0.4A

i=-i0/et/RC=-0.4/e5/0.0752(10) -6=-.4((2.303)-1/15)10(power9)A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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