When two capacitors of capacities 3- F and 6- F are connected in series and connected to 120 V- the potential difference across 3- F is-

The correct option is B 80V C=3× 69=2μ F V=120 V Q=CV=120× 2=240μ C V_3=Q3μ F=240μ C2μ F=80 V ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Parallel and Series Combination of Capacitors

When two capa...

Question

When two capacitors of capacities $3 μ F$ and $6 μ F$ are connected in series and connected to $120 V,$ the potential difference across $3 μ F$ is:

40 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

180 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $80 V$
$C = \frac{3 \times 6}{9} = 2 μ F$
$V = 120 V$
$Q = C V = 120 \times 2 = 240 μ C$
$V_{3} = \frac{Q}{3 μ F} = \frac{240 μ C}{2 μ F} = 80 V$

Suggest Corrections

Similar questions

Q. Three capacitors,

3 μ F, 6 μ F

and

6 μ F

are connected in series to a source of 120V. The potential difference, in volts, across the

3 μ F

capacitor will be

Q. Two capacitors of

3 μ F

and

6 μ F

are connected in series and a potential difference of

900 V

is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

Two capacitors of capacities 3 $μ$ F and 6 $μ F$ are connected in series and connected to 120V. The potential differences across 3 $μ$ F is $V_{0}$ and the charge here is $q_{0}$ . We have :

A) $q_{0} = 40 μ C$ B) $V_{0} = 60 V$

C) $V_{0} = 80 V$ D) $q_{0} = 240 μ C$

Three capacitors of $3 μ F, 2 μ F$ and $6 μ F$ are connected in series. When a battery of 10V is connected to this combination then charge on $3 μ F$ capacitor will be :

Q. The equivalent capacitance of capacitors

6 μ F

and

3 μ F

connected in series is ______.

Join BYJU'S Learning Program

When two capacitors of capacities 3μF and 6μF are connected in series and connected to 120V, the potential difference across 3μF is:

The correct option is B 80VC=3×69=2μFV=120VQ=CV=120×2=240μCV3=Q3μF=240μC2μF=80V

When two capacitors of capacities $3 μ F$ and $6 μ F$ are connected in series and connected to $120 V,$ the potential difference across $3 μ F$ is:

The correct option is B $80 V$
$C = \frac{3 \times 6}{9} = 2 μ F$
$V = 120 V$
$Q = C V = 120 \times 2 = 240 μ C$
$V_{3} = \frac{Q}{3 μ F} = \frac{240 μ C}{2 μ F} = 80 V$