I need help with #9, #10, #11, #12. in 10 mins thank you. 9. A series RC cirouit

Question

I need help with #9, #10, #11, #12.
in 10 mins thank you.

9. A series RC cirouit has a tinse costant of 4.8 The bathery las voltage of 14 V and the maxinmum curent jst sftur closing is 24 A. The capacitor is initially after closing the switch What is the charge on itor is initially uncharged. l the capacitor 5 s after the switch is closed? y uncharged. 50 V A) 1.34 C B) 37.27C C) 72 uC D) 24 c ) 20.16C 10. If the size of the charge value bi changst 4. 5X for bat ef ce and i nin 1.5 mutual force between them will be changed by what factor? A) 3.375 B) 3.24 C) 0.679 D) 4545 E) 1.04535 | F:ni. ( hooked in eeries with a S resistor and a 12 battery. What i. thn current in the sa A) 1.33 A B) 4.167A C) 0.625 A D) 24 A E) IA 11. Two resistors of values 6 and 12 sre oonnected Ln parallel. Thin oorhination in turn is resistance of 5.5 ohms. What is the velocity of the resistance of S,.S ohms. What is the velocity of the cart when the cunet in the rails is 0.581818 A und the field of 1.5 Tesla is pointing vertically? 2. An all metal cart is onApair ofails 0.5 m apart. The rails are connected to a 8 volt battery and have 2. An all metal cart is on apair of rails 0.5 rm

Explanation / Answer

9. Time constant = RC = 4.8s

Just after closing the switch capacitor will behave as short circuited. So equation V=IR can be applied

=> 14 = 24*R

=> R = 0.583ohm

therefore C = 4.8/R = 8.23F

we know that Q = CV = 8.23*14(1 - e(-t/4.8))

So by putting t = 5s in above equation we can get the amount of charge stored in capacitor after 5 sec

=> Q = 115.22(1 - e(-5/4.8))    72C

10. We know that mutual force on two point charges q1 and q2 separated by a distance d is given by F = kq1q2/d2   So we concude from the above equation that mutual force is directly proportional to both charges  q1 and q2 and inversely proportional to d2

Therefore the change in mutual force between them will be by a factor of 4.5*4.5/2.52 = 3.24

11. Since 5ohm resistor is in series with battery so the current in it will be the same as the overall current flowing through battery. Therefore all we have to do is minimise the given sircuit and solve for overall current flowin through circuit and that will be the answer.

Equivalent resistance of 6||12 will be 4ohm

now 4ohm is in series with 5ohm so total equivalent resistance will be 5+4=9ohm

So overall current flowing through battery with potential 12V and thus flowing through the 5ohm resistance will be I=V/R

=> I = 12/9 = 1.33A.

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