Two capacitors of capacities 3mu F and 6mu F are connected in series and connected to 120V- The potential differences across 3mu F is V-0- and the charge here is q-0- We have -A-q-0-40mu C B- V-0-60VC-V-0-80V D-q-0-240mu CA- C are correctA- B are correctB- D are correctC- D are correct

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When capacitor are connected is series charge present on capacitor is same-Now- q0-CV-240-xD7-10-x2212-6We know- V0-q0CV0-240-xD7-10-x2212-63-xD7-10-x2212-6V0-80V

Two capacitors of capacities 3μF and 6μF are connected in series and connected to 120V. The potential differences across 3μF is V0 and the charge here is q0. We have :A)q0=40μC B) V0=60VC)V0=80V D)q0=240μCA, C are correctA, B are correctB, D are correctC, D are correct

When capacitor are connected is series charge present on capacitor is same.Now, q0=CV=240×10−6We know, V0=q0CV0=240×10−63×10−6V0=80V

When capacitor are connected is series charge present on capacitor is same.
Now,
$q_{0} = C V = 240 \times 10^{- 6}$
We know,
$V_{0} = \frac{q_{0}}{C}$
$V_{0} = \frac{240 \times 10^{- 6}}{3 \times 10^{- 6}}$
$V_{0} = 80 V$