What charges will flow through section B of the circuit in the direction shown when switch S is closed

When the switch S is open, as shown in the figure(a): The net potential difference across circuit: V = V 1 + V 2 = 30 V + 60 V = 90 V . Since B is open circuited, the capacitors 2 μF and 3 μF are in series. Thus, the equivalent capacitance of the circuit, when the switch S is open is, C e q = C 1 × C 2 C 1 + C 2 = 2 × 3 5 = 6 5 μF Hence, charge in each capacitor, q = C e q V = 6 5 × 90 = 108 μC . When the switch S is closed, as shown in the figure (b): Let's take q 1 and q 2 are the charges stored in the two capacitors. Now, q 1 = C 1 V 1 = 2 × 30 = 60 μC . q 2 = C 2 V 2 = 3 × 60 = 180 μC . Now, After closing the switch, the charge flown out of the 2 μF capacitor is, ∆ q 1 = 108 - 60 = 48 μC and The charge flown into the 3 μF capacitor after the closing the switch is, ∆ q 2 = 180 - 108 = 72 μC . So the net charge flown through the section B or through the switch is, ∆ q = ∆ q 1 + ∆ q 2 = 48 + 72 = 120 μC .

The capacitor each having capacitance C = 2 μ F are connected with a battery of emf 30 V as shown in the figure. When the switch S is closed, find: (a) the amount of charge flown through the battery (b) the energy supplied by the battery (c) the heat generated in the circuit (d) the amount of charge flown through the switch S

a We can see from the figure, before closing the switch, the capacitors are connected in parallel then series it means, C ' = C + C = 2 C and C eq = C ' × C C ' + C = 2 3 C = 2 3 × 2 = 4 3 μ F ∴ q 1 = C eq V = 4 3 × 30 = 40 μC The energy stored is U i = 1 2 C eq V 2 = 6 × 10 - 4 J , After closing the switch, the capacitors which are connected in parallel say, C ' short-circuited and net capacitance becomes, C ' eq = C . So, charged stored after short-circuited, ∴ Q = C V = 2 × 30 = 60 μC , It means the change in charge stored, ∆ q = q f - q i = 60 - 40 = 20 μC Therefore, the charge flown from the battery is 20 μC And U f = 1 2 C eq ' V 2 = 1 2 C V 2 = 9 × 10 - 4 J b The energy supplied by the battery, W cell = V ∆ q = 30 × 20 × 10 - 6 = 6 × 10 - 4 = 0 .6 mJ W cell = ∆ U + ∆ H ∴ ∆ U = U f - U i = 6 × 10 - 4 - 3 × 10 - 4 = 3 × 10 - 4 J = 0 .3 mJ c Substituting the values in the above equation we get the heat generated in the circuit, ∆ H = 0 . 6 - 0 . 3 = 0 . 3 mJ d The charge flows through switch = the charge supplied by the battery + the charges flowing for neutralization of leftward capacitors = 20 + 20 + 20 = 60 μC { ∴ Charges on each leftward capacitors are 20 μC before short-circuiting}

In figure, the system is in steady state. Then (a) V A - V B = _____ (b) V B - V C = _____ (c) V D - V E = _____ (d) The energy stored in the circuit is ______

Both 3 μF capacitors will be in parallel, and similarly upper two 1 μF capacitors will be in parallel. The circuit may be redrawn as follows. Equivalent of two parallel capacitors between A and B is given as C AB = 3 + 3 = 6 μF Equivalent of two parallel capacitors between B and C is given as C BC = 1 + 1 = 2 μF Now the equivalent circuit will look like as shown below. Equivalent of series resistors in branch A C is given as C AC = 6 × 2 6 + 2 = 3 2 μF . Since, branch A C and D E are parallel to battery. Charge in branch A C is given as q 1 = 100 3 2 = 150 μC . Similarly, charge in branch D E is given as q 2 = 100 1 = 100 μC Now charge on 6 μF and 2 μF is 150 μC and charge on 1 μF is 100 μC . Potential difference between A and B is V A - V B = q 1 6 = 150 6 = 25 V Potential difference between A and C is V A - V C = q 1 3 2 = 150 3 2 = 75 V Potential difference between D and E is V D - V E = q 2 1 = 100 1 = 100 V Equivalent capacitance of the circuit is given as C eq = C AC + C DE = 3 2 + 1 = 5 2 μF . Energy stored in the capacitor is given as U = 1 2 C eq V 2 = 1 2 5 2 × 10 - 6 × 100 2 ⇒ U = 125 × 10 - 4 J = 12 . 5 mJ

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Four capacitors are connected as shown in the figure to a $30 V$ battery. Find the potential difference between points $a$ and $b$.

Important Questions on Capacitor and Capacitance

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JEE Advanced

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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 8

In figure, the battery has a potential difference of $20 V$ . Find

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(a) the equivalent capacitance of all the capacitors across the battery and

(b) the charge stored on that equivalent capacitance.

Find the charge on

(d) capacitor $C_{2}$ , and

(e) capacitor $C_{3}$ .

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JEE Advanced

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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 9

What charges will flow through section $C_{3} = 2.5 μF$ of the circuit in the direction shown when switch $S$ is closed
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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 10

The figure shows a network of seven capacitors. If charge on $5 μF$ capacitor is $10 μC$ , find the potential difference between points $A$ and $C$.
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JEE Advanced

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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 11

The capacitor each having capacitance $C = 2 μ F$ are connected with a battery of emf $30 V$ as shown in the figure. When the switch $S$ is closed, find:

Question Image

(a) the amount of charge flown through the battery

(b) the energy supplied by the battery

(d) the amount of charge flown through the switch $S$

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JEE Advanced

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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 1

For the network of capacitors as shown in figure.

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(a) Find the potential of the junction $B,$

(b) Find the potential of the junction $D,$

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JEE Advanced

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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 2

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In figure, the system is in steady state. Then

(a) $V_{A} - V_{B} =$ _____

(b) $V_{B} - V_{C} =$ _____

(d) The energy stored in the circuit is ______

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JEE Advanced

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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 3

A $10 μ F$ capacitor is charged to $15 V$ . It is next connected in series with an uncharged $5 μ F$ capacitor. The series combination is finally connected across a $30 V$ battery, as shown in figure. Find the new potential difference across the $5 μ F$ and $10 μ F$ capacitors.
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JEE Advanced

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PHYSICS for Joint Entrance Examination JEE (Advanced) Electrostatics and Current Electricity>Chapter 4 - Capacitor and Capacitance>Concept Application Exercise>Q 4

Two capacitors $C_{1} = C_{2} = C$ are connected with a battery of emf $ε$ as shown in figure. The switch $S$ is open for a long time and then closed.

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(a) Find the charge flowing through the battery when the switch $S$ is closed.

(b) Find the work done by the battery.

(d) Find the heat developed in the system.

Four capacitors are connected as shown in the figure to a 30 V battery. Find the potential difference between points $a$ and $b$.

Important Questions on Capacitor and Capacitance

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Four capacitors are connected as shown in the figure to a $30 V$ battery. Find the potential difference between points $a$ and $b$.