Two identical capacitors have the same capacitance C. One of them is charged to potential $V_{1}$ and the other to $V_{2}$ . The capacitors are connected after charging. The decrease in energy of the combined system is :

\frac{1}{4} C (V_{1}^{2} - V_{2}^{2})

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

\frac{1}{4} C (V_{1}^{2} + V_{2}^{2})

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

\frac{1}{4} C {(V_{1} - V_{2})}_{1}^{2}

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

\frac{1}{4} C {(V_{1} + V_{2})}_{1}^{2}

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

Open in App

Solution

The correct option is D $\frac{1}{4} C {(V_{1} - V_{2})}_{1}^{2}$
When connected total charge
$Q = C (V_{1} + V_{2})$
when connected in parallel
$\Rightarrow x = C \cdot V$
$Q - x = C \cdot V .$
$\frac{x}{Q - x} = 1$
$x = \frac{Q}{2}$
$\Rightarrow V = \frac{Q}{2 C}$
$\Rightarrow V = \frac{V_{1} + V_{2}}{2}$
$∴ d e c r e a s e i n e n e r g y = \frac{1}{2} C V_{1}^{2} + \frac{1}{2} C V_{2}^{2} - \frac{1}{c} \cdot C \cdot (\frac{V_{1} + V_{2}}{2})^{2}$
$Δ E = \frac{1}{4} C (V_{1} - V_{2})^{2}$

Suggest Corrections

Similar questions

Q. Two identical capacitors, have the same capacitance

C

. One of them is charged to potential

V_{1}

and the other to

V_{2}

. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is:

Q. Two identical capacitors have the same capacitance C. One of them is charged to potential

V_{1}

and the other to

V_{2}

. The negative ends of the capacitors are connected together . When the positive ends are also connected, the decrease in energy of the combine system is

Two identical photo-cathodes receive light of frequencies $f_{1} a n d f_{2}$ . If the velocities of the photo electrons (of mass m) coming out are respectively $v_{1} a n d v_{2}$ , then

Q. Two identical, photocathodes receive light of frequencies

f_{1}

and

f_{2}

. If the velocities of the photo electrons (of mass m) coming out are respectively

v_{1}

and

v_{2}

, then (where the symbols have their usual meaning)

If instantaneous voltage across R is

V_{1}

& across C is

V_{2}

, then total voltage drop across R & C at same instant will be

Join BYJU'S Learning Program

Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The capacitors are connected after charging. The decrease in energy of the combined system is :

The correct option is D 14C(V1−V2)2When connected total chargeQ=C(V1+V2)when connected in parallel⇒x=C⋅VQ−x=C⋅V.xQ−x=1x=Q2⇒V=Q2C⇒V=V1+V22∴decrease in energy=12CV21+12CV22−1c⋅C⋅(V1+V22)2ΔE=14C(V1−V2)2

Two identical capacitors have the same capacitance C. One of them is charged to potential $V_{1}$ and the other to $V_{2}$ . The capacitors are connected after charging. The decrease in energy of the combined system is :