The figure shows a diagonal symmetric arrangement of capacitors and a battery
If the potential of C is zero, then
The capacitance of a metallic sphere will be 1mF, if its radius is nearly
The expression of capacitance of metallic sphere is C=4πϵ_{0}r (The capacitance of a single spherical conductor means that we imagine a spherical shell of infinite radius surrounding the first conductor.)
Therefore,
r = C/4πϵ_{0}
= 10 ^{− 6} × 9 × 10^{9} = 9 × 10^{3} m = 9km
Two capacitors of capacitances 1mF and 3mF are charged to the same voltages 5V. They are connected in parallel with oppositely charged plates connected together. Then
Two spherical conductors A and B of radii R and 2R respectively are each given a charge Q. When they are connected by a metallic wire. The charge will
The capacity of a parallel plate condenser is C. Its capacity when the separation between the plates is halved will be
The capacity of a parallel plate condenser is given as,
C=εA/d
When the separation between the plates is halved then the new capacity of the condenser is given as,
C′=εA/(d/2)
C′=2C
Thus, the new capacity of parallel plate condenser is 2C.
If the p.d. across the ends of a capacitor 4mF is 1.0 kilovolt. Then its electrical potential energy will be
The energy of a charged capacitor resides in
The energy of a charged capacitor resides in both electric and magnetic fields .
Energy resides in the electric field because of the charges on the capacitor.
Energy resides in the magnetic field because of The Maxwell's displacement current in the capacitor.
The capacitance of a parallel plate condenser does not depend upon
The capacitance C is the amount of charge stored per volt, or C=QV. The capacitance of a parallel plate capacitor is C=ϵ_{0}Ad, when the plates are separated by air or free space.
The energy density in a parallel plate capacitor is given as 2.2 × 10^{10} J/m^{3}. The value of the electric field in the region between the plates is 
Two capacitances of capacity C_{1} and C_{2} are connected in series and potential difference V is applied across it. Then the potential difference acros C_{1} will be
A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearly
Given,
C_{1}=0.5 mf
C_{2}=0.2mf
V=100v
So,
U=(1/2) C_{T}V^{2}
C_{T}=C_{1}C_{2}/XC_{1}+C_{2}=(0.5x0.2/0.5+0.2)mf
C_{T}=0.14mf
U=(1/2)(0.14mf)(100V)^{2}
=(10.07x1^{6}x10^{4})J
=(7x10^{2}x10^{6}x10^{4})J
=7x10^{4}J
Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential is
First find the charge in 1st capacitor
Q_{1}= C_{1}xV_{1}
=0.003*300
=0.9 C
Similarly, find Q_{2}
i.e. Q_{2}=2.5 C
After connecting the capacitors in parallel you get a net capacitance C=8 mF (c_{1}+c_{2}).
Thus, final potential is V=Q/C
Q=Q_{1}+Q_{2}=3.4 C
V=3.4/0.008
= 425 V
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connect in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
Where N is the number of turns of the toroid coil, I is the amount of current flowing and r is the radius of the toroid.
As both capacitors are in parallel so equivalent capacitance C_{eq}=C+2C=3C .
Net potential, V_{n}=2V−V=V
Energy stored, U=(1/2)C_{eq}V_{n}^{2}=(1/2)×3CV^{2}=(3/2)CV^{2}
A 2μF capacitor is charged to a potential = 10 V. Another 4μF capacitor is charged to a potential = 20V. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. What heat is evolved in the circuit?
Charge on 2μf capacitor=10vx2μF=20μC=Q_{1}
Charge on 4μf capacitor=4μFx20v=80μC=Q_{2}
Now, these capacitors are connected in a single loop in given manner,
Now after transfer,
Applying loop rule,
(q/2)+[(60+q)/4]=0=>q=20μC
Now, drop across the capacitors is,
ΔV_{f}=40/4v=10v
Final energy stored in capacitor=U_{f}
= (1/2) (C1+C2)(ΔVf)^{2}=1/2 x6 x100uf=300 μf
Initial energy stored in capacitor=U_{i}
= (1/2) C_{1}V_{1}^{2}+(1/2) C_{2}V_{2}^{2}=(1/2)x2x100+ (1/2)x4x400=900 μf
Now, heat evolved in the circuit will be,
U_{i}U_{f}= (900300) μf=600 μf
Thus, the correct answer would be option B.
The figure shows a diagonal symmetric arrangement of capacitors and a battery
Identify the correct statements.
If 2μF where not present between B and D, then potential drop across upper 4μF will be less than potential drop across lower 2μF,i.e.,
In the circuit shown in figure charge stored in the capacitor of capacity 5μf is
Plate A of a parallel air filled capacitor is connected to a spring having force constant k and plate B is fixed. If a charge +q is placed on plate A and charge _q on plate B then find out extension in spring in equilibrim. Assume area of plate is `A'
At equilibrium,
F_{electric}=F_{spring}
qE=kx ____(1)
we know that,
C=Q/V
And V=ε.d
C=Aε0/d
Therefore, ε=θ/Aε_{0}
Here, ε=q/ Aε0
Putting the value in the equation (1)
q/ Aε_{0}=kx
x=q^{2}/akε_{0}
Three uncharged capacitors of capacitane C_{1} = 1mF, C_{2} = 2mF and C_{3} = 3mF are connected as shown in figure to one another and to points A, B and D potential f_{A} = 10V, f_{B} = 25V and f_{D} = 20 V, Determine the potential (f_{0}) at point O.
⇒q_{1}+q_{2}+q_{3}=0
⇒C_{1}(V_{o}−V_{A})+C_{2}(V_{o}−V_{B})+C_{3}(V_{o}−V_{D})=0
⇒(C_{1}+C_{2}+C_{3})Vo=C_{1}V_{A}+C_{2}V_{B}+C_{3}V_{D}
V_{o}= C_{1}V_{A}+C_{2}V_{B}+C_{3}V_{D}/ C_{1}+C_{2}+C_{3}
= [(1×10)+(2×25)+(3×20)]/(1+2+3)
=(10+50+60)/6
=20Volts
Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between the points M and N.
Five capacitors 4μF, 2μF, 4μF, 6μF and 1.2μF are connected as shown in the figure.
Here, Equivalent capacitance = ½,
EMF = 24
Thus, charge passes through each capacitor = 12 μC
Voltage across 4μF = 3V
Voltage across 6μF = 2V
=> Potential difference (V):
V = 3 + 7+ 2 = 12V
Thus, potential difference (V) between the two given points M and N is 12v.
Find the potential difference V_{a}  V_{b} between the points a and b shows in each parts of the figure.
a.In this figure, the left and the right branch is symmetry. So the current go to the branch 'ab' for both sides are opposite and equal. Hence it cancels out.
Thus, net charge, Q= 0. ∴V=Q/C=0/C=0
∴Vab=0
b.The net potential, V= Net charge /Net capacitance =C1V1+C2V2+C3V3/7
V=(24+24+24)/7=72/7=10.28V
Each plate of a parallel plate air capacitor has an area S. What amount of work has to be performed to slowly increase the distance between the plates from x_{1} to x_{2} If the charge of the capacitor, which is equal to q or
Given,
Area of plateS
Initial distance=X_{1}
Final distance=X_{2}
Charge given, q
Initial capacitance
C_{1}=Sε/ X_{1}
Energy stored
V_{1}=q^{2}/2C=q^{2}/2(Sε/X_{1})
U_{1}= q^{2}X_{1}/2Sε
Final capacitance
C_{2}= Sε/X_{2}
Energy stored,
U_{2}= q^{2} x_{2}/2Sε
Amount of work done=Change in stored energy between capacitor
=U_{f}U_{i}=U_{2}U1
=(q^{2}X_{2}/2Sε) (q^{2}X_{1}/2Sε)
Work done=q^{2}/2Sε (X_{2}X_{1})
Each plate of a parallel plate air capacitor has an area S. What amount of work has to be performed to slowly increase the distance between the plates from x_{1} to x_{2} If the voltage across the capacitor, which is equal to V, is kept constant in the process.
When voltage is kept const., the force acing on each plate of capacitor will depend on the distance between the plates.
From energy conversion,
U_{f}U_{i}=A_{cell}+A_{agent}
Or,(1/2)(ε_{0}S/x2)V2(1/2)(ε_{0}S/x1)V^{2}
=[(ε_{0}S/x_{2}) (ε_{0}S/x1)] V2 +A_{agent}
(as A_{cell}=(q_{f}q_{1})V=(CfCi)V2)
So, A_{agent}= (ε_{0}SV^{2}/2)[(1/x_{1}))1/x_{2})]
If charge on left plane of the 5F capacitor in the circuit segment shown in the figure is20C, the charge on the right plate of 3F capacitor is
If the charge on left plate of 5μF is −20μC, then the charge on right plate of 5μF is +20μC
So due to the polarization the charge on the left plate of 3μF is negative and the charge on the right plate of 3μF is positive.
thus, charge on right plate of 3μF is Q=[3/(3+4)]×20=60/7=8.57μC
In the circuit shown, the energy stored in 1μF capacitor is
The capacitance 4 and (3,5,1) are in parallel so potential across (3,5,1) will be V=24V.
The equivalent capacitance of (3,5,1) is C_{eq}=3(5+1)/{3+(5+1)}=2μF
and Q_{eq}=C_{eq}V=2×24=48μC
As 3 and (5,1) are in series so charge on 3 and (5,1) is equal to Q_{eq}.
Thus potential across (5,1) is V_{51}= Q_{eq}/C_{51}=48/(5+1)=8V
As 5 and 1 are in parallel so potential difference of capacitors will be same i.e 8V
Now energy stored in 1 is =(1/2)C_{1}V_{51}^{2}=(1/2)×10^{−6}×82=32μJ
A capacitor C_{1} = 4mF is connected in series with another capacitor C_{2} = 1mF. The combination is connected across a d.c. source of voltage 200V. The ratio of potential across C_{1} and C_{2} is
Voltage in series connections are,
V_{1}=C_{2}/{(C_{1}+C_{2}) x V} V_{2}={C_{1}/(C_{1}+C_{2})xV}]
V_{1}=(1/S) x 200
V_{2}=(4/S) x200
Ration, V_{1}/V_{2}=((1/S) x200)/((4/S)x200)=1:4
A 5.0 mF capacitor having a charge of 20 mC is discharged through a wire of resistance of 5.0 W. Find the heat dissipated in the wire between 25 to 50 ms after the capactions are made.
A charged capacitor is allowed to discharge through a resistance 2Ω by closing the switch S at the instant t = 0. At time t = ln2μs, the reading of the ammeter falls half of its initial value. The resistance of the ammeter equal to
A capacitor C =100μF is connected to three resistor each of resistance 1 kW and a battery of emf 9V. The switch S has been closed for long time so as to charge the capacitor. When switch S is opened, the capacitor. Discharges with time constant
In the transient shown the time constant of the circuit is :
Applying Thevenin's theorem we consider the circuit from right side and short circuiting the voltage sources and opening the extreme right resistance we get the equivalent resistance
=1/Rth=(1/3R)+(1/R)=4/3R=3R/4 Rth=Thevenin’s resistance
So now adding the remaining resistance R with Rth in series we get the equivalent resistance
Req=(3R/4)+R=7R/4
And time constant τ=RC
τ=(7/4)RC
In the circuit shown in figure C_{1} =2C_{2 }. Switch S is closed at time t=0. Let i_{1} and i_{2} be the currents flowing through C_{1} and C_{2} at any time t, then the ratio i_{1 }/ i_{2 }
The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.
The answer provided in the forum is wrong. the correct answer is option A.B,C.
Solution:
As the emf of cell is ϵ=Q/C, charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor.
As the Y is connected to the negative terminal of the cell so it is grounded and the total charge on plate Y will be zero.
Here cell charge the capacitor plate X and its total charge =(Q+Q)=2Q
The energy will supply by cell is U=(½)Cϵ2
The correct option is A B C only.
Two capacitors of capacitances 1mF and 3mF are charged to the same voltages 5V. They are connected in parallel with oppositely charged plates connected together. Then
V=(C_{1}V_{1}−C_{2}V_{2})/C_{1}+C_{2}
=3×5−1×5/1+3
= 2.5volt
Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,
When a parallel plates capacitor is connected to a source of constant potential difference,
When two identical capacitors are charged individually to different potentials and connected parallel to each other, after disconnecting them from the source
Two thin conducting shells of radii R and 3R are shown in the figure. The outer shell carries a charge +Q and the inner shell is neutral. The inner shell is earthed with the help of a switch S.
(A) Potential of inner shell is due to potential developed by outer shell.
(B) Since the inner shell is Earthed so its potential becomes zero.
(C) V_{inner}=(KQ/3R)+(KQI/R)=0⇒rQ'=−Q/3
(D) It capacitance becomes =4π∈_{0}R_{1}R_{2}/R_{2}−R_{1}+4π∈R_{2}
Therefore the correct answer is option A B C D.
Two capacitors of 2mF and 3mF are charged to 150 volt and 120 volt respectively. The plates of capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5 mF falls to the free ends of the wire. Then
Charge in 2μF Capacitor =300μC
Charge in 3μF Capacitor =360μC
when 1.5μF falls on the circuit ends, Redistribution of charge now will be:
Charge in 2μF Capacitor =300−xμC
Charge in 3μF Capacitor =360−xμC
Charge in 1.5μF Capacitor =xμC
By Kirchoff Voltage Law:
x/1.5=[(300−x)/2]+[(360−x)/3]
which gives x=180μC
Therefore, Answers are A, B & C
In the circuit shown initially C_{1}, C_{2} are uncharged. After closing the switch
Here C_{1} and C_{2} are in series and the total potential between the capacitors is V_{T}=12+6=18V
C_{eq}=(4×8)/(4+8)=(8/3) μF and
Q_{eq}=C_{eq}V_{T}=(8/3)×18=48μC
As they are in series so the charge on each capacitor is equal to Q_{eq}=48μC.
∴Q1=Q2=48μC
Potential across C_{1} is V_{1}= Q_{1}/C_{1} =48/4=12V
Potential across C_{2} is V_{2}= Q^{2}/C_{2}=48/8=6V
A circuit shown in the figure consists of a battery of emf 10V and two capacitance C_{1 }and C_{2} of capacitances 1.0mF and 2.0mF respectively. The potential difference V_{A} ^{_} V_{B} is 5V
Here the capacitors are in series so the charge on each is equal to equivalent charge.
The equivalent capacitance is Ceq=(1×2)/(1+2) =(2/3)μF and the equivalent charge , Q_{eq}=C_{eq}V=(2/3)(10+5)=10μC
Thus, Q_{1}=Q_{2}=Q_{eq}=10μC
Potential across C1 is V1=Q1/C1=10/1=10V and Potential across C_{2} is V_{2}=Q_{2}/C_{2}=10/2=5V
Two capacitors of equal capacitance (C_{1} = C_{2}) are shown in the figure. Initially, while the switch S is open, one of the capacitors is uncharged and the other carries charge Q_{0}. The energy stored in the charged capacitor is U_{0}. Sometimes after the switch is colsed, the capacitors C_{1} and C_{2} carry charges Q_{1} and Q_{2}, respectively, the voltages across the capacitors are V_{1} and V_{2}, and the energies stored in the capacitors are U_{1} and U_{2}. Which of the following statements is INCORRECT ?
In the above diagram, C_{1} and C_{2} are connected in parallel.
Hence, both get the same voltage. ∴V_{1}=V_{2} is correct.
We know that Q_{1}=C_{1}V_{1} and Q_{2}=C_{2}V_{2}
∵C_{1}=C_{2}, Q_{1}=Q_{2}
We know that energy U_{1}=(1/2)C_{1}V_{1}^{2} and U_{2}=(1/2)C2V_{2}^{2}
By closing the switch, the C_{1} or V_{1} doesn't get affected.
Hence U_{0}=U_{1} and not U_{1}+U_{2}
Similarly, by closing the switch, the Q_{1} or V_{1} doesn't get affected.
Hence Q_{0}=Q_{1}
∵Q_{0}=Q_{1}=Q_{2}, Q_{0}=(1/2)(Q_{1}+Q_{2})
The figure shows a diagonal symmetric arrangement of capacitors and a battery
Identify the correct statements.
Now V_{A}−V_{B}<V_{A}−V_{D}⇒V_{B}>V_{D}
Now if 2μF is connected between B and D, charge will flow from B and D and Finally VB>VD.
V_{A}=20V. At junction B,
q_{2}=q_{1}+q_{3}
or (V_{A}−V_{B})=(V_{B}−V_{D})2+(V_{B}−V_{C})2
or 4(V_{A}−V_{B})+2(V_{D}−V_{B})=2V_{B}
At junction D:
q_{2}=q_{1}+q3
or 4(V_{D}−V_{C})=2(V_{A}−V_{D})+2(V_{B}−V_{D})
or 2(V_{A}−V_{D})+2(V,−V_{D})=4V_{D}.
The figure shows a diagonal symmetric arrangement of capacitors and a battery
If the potential of C is zero, then
For very low frequencies, capacitor acts as _______
Capacitive impedance is inversely proportional to frequency. Hence at very low frequencies the impedance is almost infinity and hence acts as an open circuit and no current flows through it.
The charge across the capacitor in two different RC circuits 1 and 2 are plotted as shown in figure.
Choose the correct statement (s) related to the two circuits
The charge on capacitor at any time t is q=q_{max}(1−e^{−t/CR})
As the q_{max} of both curves is the same so both capacitors will be charged to the same charge.
As (2) curves take more time for maximum charging so the time constant of curve (2) is greater than (1). i.e, τ2>τ_{1}⇒C_{2}R_{2}>C_{1}R_{1}
As qmax is same for both curve so C_{1}E_{1}=C_{2}E_{2}
Since C_{1} and C_{2} are different so emf of both cells may be different.
now E_{1}=(C_{2}/C_{1})E_{2}>(R_{1}/R_{2})E_{2}
If R_{1}>R_{2} , then E_{1}>E_{2}.
The charge across the capacitor in two different RC circuits 1 and 2 are plotted as shown in figure.
Identify the correct statement(s) related to the R_{1}, R_{2}, C_{1} and C_{2} of the two RC circuits.
The charge on capacitor at any time t is q=q_{max}(1−e^{−t/CR})
As the qmax of both curves is the same so both capacitors will be charged to the same charge.
As (2) curve take more time for maximum charging so the time constant of curve (2) is greater than (1). i.e, τ_{2}>τ_{1}⇒C_{2}R_{2}>C_{1}R_{1}
As qmax is same for both curve so C_{1}E_{1}=C_{2}E_{2}
Since C_{1} and C_{2} are different so emf of both cells may be different.
now E_{1}=(C_{2}/C_{1})E_{2}>(R_{1}/R_{2})E_{2}
If R_{1}>R_{2} , then E_{1}>E_{2}.






