Question

A capacitor of unknown capacitance is connected across a battery of V volts.The charge stored in it is 360 MicroCOulomb.When Potential across the capacitor is reduced by 120V,the charge stored in it becomes 120 micro coulomb. Calculate (i) the potential V and the unknown capacitance ; (ii) what will be the charge stored in the capacitor,if the voltage applied had increased by 120 V?

Answer 1

Hey !!

(1) Let the initial voltage = V volts

Charge stored Q₁ = 360 μC

                Q₁ = CV -----------------> (A)

Now the changed potential

              V₂ = V - 120 volts

Charged stored Q₂ = 120 μC

                          Q₂ = CV₂        ----------------> (B)

Now divide equation B from A, we have

            Q₁ / Q₂ = CV₁ / CV₂

360 μC / 120 μC = V / V - 120

On solving we get V = 180 volts

Now the value of unknown capacitance C can be calculate as

                       C = Q₁ / V

                     = 360 × 10⁻⁶ / 180

(2) When the voltage applied increases to 120 V, then

                        V₃ = CV₃

finally the charge stored in the capacitor will be,

                  Q₃ = CV₃

                            = 2 × 10⁻⁶ × 300

                          = 600 μC

GOOD DAY !!

Answer 2

Answer:

very simple answer

1. Q=CV

CV= 360 MICRO COULOMB.............. (1)

on reducing v by120 volt

Q' = C(v-120)

C(v-120) = 120 micro coulomb......... (2)

remaining answer is in the picture