Question

11. Six 2-ohm resistors are placed in series and connected across a 12-volt battery’s terminals. The voltage I cannot easily obtain from this system is

(a) 3V

(b) 6V

(c) 9V

(d) 12V

Two 2 microfarad capacitors are connected in series and the combination placed across the terminals of a 12 V battery. What charge will appear on their positive plates?

(a) 1 μC

(b) 2 μC

(c) 4 μC

(d) 12 μC

Can someone explain these questions

Answer 1

remember these about Combination of capacitors

if C1 and C2 are the individual capacitor are connected in series,

then their resultant capaciatnce 1/Cnet = 1/C1+1/C2 or Cnet = C1C2/(C1+C2)

if c1 and C2 are connected in parallel, then their effective capacitancen is Cnet = C1+C2

also in sereis combination same amount of charge passes through each of the capacitiors

and in parallel combination, same PD (voltage) exists across each of the capacitor

remember these about Combination of Resistors

if R1 and R2 are the resistances connected in series then effective resiatcne Rnet = R1+R2

if R1 and R2 are connected in parallel, then 1/Rnet =1/R1 +1/R2 or Rnet = R1R2/(R1+R2)

same current passes through each of the ressitor in series combination and

same votage passes through eacf of the resistor in parallel combination

apply ohms law V = iR

where V = volatge and i = current

so here

11. Rnet of 6 two ohms resistors in series = Rs = nR = 6*2 = 12 ohms

using ohms law V = iR

i = 12/12 = 1Amps

Voltage across resisotr 1 = 1*2 = 42

Volatge across ressitor 2 = 1* 4 = 4 Volts

Volatge across ressitor3 = 1* 6 = 6 Volts

Volatge across ressitor 4 = 1* 8 = 8 Volts

Volatge across ressitor 5 = 1* 10 = 10Volts   

Volatge across ressitor 2 = 1* 12 = 12 Volts

so answers are option A and C ( 3 and9 Volts)

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in series, 1/Cnet = 1/C1+1/C2

1/Cnet = (1/2) + 1/2 = 1

Cnet = 1uF

Qnet = Q1 = Q2 = CV = 12 * 1 = 12uC

so answer is option d , 12 uC