JEE Advanced PYQ Physics – Miscellaneous Mixed Concept Problems

Q 1 :

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes $q_{1} μ C$ . Then S is switched to position Q. After a long time, the charge on the capacitor is $q_{2} μ C$ .

[IMAGE 1000]

Q. The magnitude of $q_{1}$ is ______. [2021]

(1.33)

$When the switch is connected to position P$

[IMAGE 1001]

From KVL,

$V_{A} - 1 \cdot i_{1} - 1 + 2 - 2 i_{1} = V_{A} \Rightarrow 3 i_{1} = 1$ $∴$ $i_{1} = \frac{1}{3} A$

Again, $V_{A} - 1 \cdot i_{1} - 1 = V_{B}$ or, $V_{A} - V_{B} = 1 + i_{1} = \frac{4}{3} V$

Potential drop across the capacitor, $Δ V = \frac{4}{3} V$

$∴$ $Charge on capacitor, q_{1} = C Δ V = 1 \times \frac{4}{3} μ C$

$q_{1} = 1.33 μ C$

$When the switch is connected to position Q$

[IMAGE 1002]

From KVL,

$V_{A} - 1 \cdot i_{2} + 2 - 2 i_{2} = V_{A}$ $\Rightarrow 3 i_{2} = 2$ $∴$ $i_{2} = \frac{2}{3} A$

Again, $V_{A} - i_{2} \times 1 = V_{B}$

$V_{A} - V_{B} = i_{2} \times 1 = \frac{2}{3} V$

Potential difference across the capacitor, $Δ V = \frac{2}{3} V$

$∴$ $Charge on capacitor, q_{2} = C Δ V = 1 \times \frac{2}{3} μ C$

$q_{2} = 0.67 μ C$

Q 2 :

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes $q_{1} μ C$ . Then S is switched to position Q. After a long time, the charge on the capacitor is $q_{2} μ C$ .

[IMAGE 1003]

Q. The magnitude of $q_{2}$ is _____. [2021]

(0.67)

$When the switch is connected to position P$

[IMAGE 1004]

From KVL,

$V_{A} - 1 \cdot i_{1} - 1 + 2 - 2 i_{1} = V_{A}$ $\Rightarrow 3 i_{1} = 1$ $∴$ $i_{1} = \frac{1}{3} A$

Again, $V_{A} - 1 \cdot i_{1} - 1 = V_{B}$ or, $V_{A} - V_{B} = 1 + i_{1} = \frac{4}{3} V$

Potential drop across the capacitor, $Δ V = \frac{4}{3} V$

$∴$ $Charge on capacitor, q_{1} = C Δ V = 1 \times \frac{4}{3} μ C$

$q_{1} = 1.33 μ C$

$When the switch is connected to position Q$

[IMAGE 1005]

From KVL,

$V_{A} - 1 \cdot i_{2} + 2 - 2 i_{2} = V_{A}$ $\Rightarrow 3 i_{2} = 2$ $∴$ $i_{2} = \frac{2}{3} A$

Again, $V_{A} - i_{2} \times 1 = V_{B}$

$V_{A} - V_{B} = i_{2} \times 1 = \frac{2}{3} V$

Potential difference across the capacitor, $Δ V = \frac{2}{3} V$

$∴$ $Charge on capacitor, q_{2} = C Δ V = 1 \times \frac{2}{3} μ C$

$q_{2} = 0.67 μ C$

Q 3 :

A series R-C circuit is connected to an AC voltage source. Consider two cases: (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current $I_{R}$ through the resistor and voltage $V_{C}$ across the capacitor are compared in the two cases. Which of the following is/are true? [2011]

$I_{R}^{A} > I_{R}^{B}$
$I_{R}^{A} < I_{R}^{B}$
$V_{C}^{A} > V_{C}^{B}$
$V_{C}^{A} < V_{C}^{B}$

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Select one or more options

(1, 3)

In an RC-circuit, the impedance is $Z = \sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}$

The capacitance in case B is four times the capacitance in case A

$∴$ Impedance in case B is less than that of case A $(Z_{B} < Z_{A})$

Now, $I = \frac{V}{Z},$

$∴$ $I_{R}^{A} < I_{R}^{B}$

and $V_{R}^{A} < V_{R}^{B}$ $\Rightarrow V_{C}^{A} > V_{C}^{B}$

[since, if V is the applied potential difference across the series RC circuit then $V = \sqrt{V_{R}^{2} + V_{C}^{2}}$ ]

Q 4 :

A circuit with an electrical load having impedance ... is connected with an AC source as shown in the diagram. The source voltage varies in time as $V (t) = 300 \sin (400 t) V$ , where $t$ is time in seconds. List-I shows various options for the load. The possible currents $i (t)$ in the circuit as a function of time are given in List-II. [2025]

[IMAGE 1006]

Choose the option that describes the correct match between the entries in List-I and those in List-II.

[IMAGE 1007]

(P) → (3), (Q) → (5), (R) → (2), (S) → (1)
(P) → (1), (Q) → (5), (R) → (2), (S) → (3)
(P) → (3), (Q) → (4), (R) → (2), (S) → (1)
(P) → (1), (Q) → (4), (R) → (2), (S) → (5)

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(1)

$Given V (t) = 300 \sin (400 t)$

[IMAGE 1008]

$i_{0} = \frac{300}{30} = 10 A, ϕ = 0$

$So, (P) \to (3)$

[IMAGE 1009]

$Z = \sqrt{R^{2} + {(ω L)}^{2}}$

$= \sqrt{30^{2} + {(400 \times 100 \times 10^{- 3})}^{2}} = 50 Ω$

$∴$ $i_{0} = \frac{300}{50} = 6 A$

and $ϕ = \tan^{- 1} (\frac{ω L}{R}) = 53 °$

i.e., the current lags by $53 °$

$So, (Q) \to (5)$

(R) $Z = \sqrt{{(30)}^{2} + {(\frac{1}{400 \times 500 \times 10^{- 6}} - 400 \times 25 \times 10^{- 3})}^{2}}$

$= \sqrt{30^{2} + 40^{2}} = 50 Ω$

$So, (R) \to (2)$

(S) $Z = \sqrt{60^{2} + {(\frac{1}{50 \times 10^{- 6} \times 400} - 125 \times 10^{- 3} \times 400)}^{2}}$

$= \sqrt{60^{2} + {(50 - 50)}^{2}} = 60 Ω$

$∴$ $i_{0} = \frac{300}{60} = 5 A$

$So, (S) \to (1)$

Q 5 :

You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current $I$ (steady state for DC or rms for AC) flows through the circuit, the corresponding voltages $V_{1}$ and $V_{2}$ (indicated in the circuits) are related as shown in Column I. Match the two columns. [2010]

[IMAGE 1010]

A-r,s,t; B-q,r,s,t; C-p,q; D-q,r,s,t
A-q,r,s,t; B-q,r,s,t; C-p,q; D-r,s,t
A-q,r,s,t; B-p,q; C-q,r,s,t; D-r,s,t
A-p,q; B-q,r,s,t; C-q,r,s,t; D-r,s,t

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(1)

For DC circuit, in steady state, the current $I$ through the capacitor (C) is zero. In case of L-C circuit, the potential difference (V) across the inductor (L) is zero and that across the capacitor = applied potential difference. In case of L-R circuit, = (V) across the inductor (L) = across (R) = applied voltage.

For AC circuit in steady state, $I_{rms}$ current flows through the capacitor (C), inductor (R) and (L) and resistor (R). The potential difference across the resistor, inductor and capacitor $I$ . And for changing current, the potential difference across (V) inductor (L), capacitor (C), or resistor (R) is proportional to the current $(I)$ .

Q 6 :

In the given circuit, the capacitor (C) may be charged through resistance R by a battery V by closing switch $S_{1}$ . Also, when $S_{1}$ is opened and $S_{2}$ is closed, the capacitor is connected in series with inductor (L).

[IMAGE 1011]

Q. At the start, the capacitor was uncharged. When switch $S_{1}$ is closed and $S_{2}$ is kept open, the time constant of this circuit is $τ$ . Which of the following is correct? [2006]

After time interval $τ$ , charge on the capacitor is $\frac{C V}{2}$
After time interval $2 τ$ , charge on the capacitor is $C V (1 - e^{- 2})$
The work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.
After time interval $2 τ$ , charge on the capacitor is $C V (1 - e^{- 1})$

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(2)

When charging is complete, the potential difference between the capacitor plates will be V, and the charge stored in this case will be maximum.

[IMAGE 1012]

$∴$ $Q_{0} = C V$

When $t = 2 τ,$

$Q = C V [1 - e^{- \frac{2 τ}{τ}}]$

$= C V (1 - e^{- 2})$

Q 7 :

In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch $S_{1}$ . Also when $S_{1}$ is opened and $S_{2}$ is closed the capacitor is connected in series with inductor (L).

[IMAGE 1013]

Q. When the capacitor gets charged completely, $S_{1}$ is opened and $S_{2}$ is closed. Then, [2006]

at $t = 0$ , energy stored in the circuit is purely in the form of magnetic energy
at any time $t > 0$ , current in the circuit is in the same direction
at $t > 0$ , there is no exchange of energy between the inductor and capacitor
at any time $t > 0$ , instantaneous current in the circuit may be $V \sqrt{\frac{C}{L}}$

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(4)

Instantaneous charge on plates at any time $t$ during discharging

[IMAGE 1014]

$Q_{inst} = Q_{0} \cos ω t$

$∴$ Instantaneous current,

$I_{inst} = \frac{d Q}{d t} = Q_{0} ω \sin ω t$

$∴$ $I_{\max} = Q_{0} ω$

Here, $Q_{0} = C V$ and $ω = \frac{1}{\sqrt{L C}}$

$∴$ $I_{\max} = C V \times \frac{1}{\sqrt{L C}} = V \sqrt{\frac{C}{L}}$

Q 8 :

In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch $S_{1}$ . Also when $S_{1}$ is opened and $S_{2}$ is closed the capacitor is connected in series with inductor (L).

[IMAGE 1015]

Q. Given that the total charge stored in the LC circuit is $Q_{0}$ , for $t \geq 0$ , the charge on the capacitor is [2006]

$Q = Q_{0} \cos (\frac{π}{2} + \frac{t}{\sqrt{L C}})$
$Q = Q_{0} \cos (\frac{π}{2} - \frac{t}{\sqrt{L C}})$
$Q = - L C \frac{d^{2} Q}{d t^{2}}$
$Q = - \frac{1}{\sqrt{L C}} \frac{d^{2} Q}{d t^{2}}$

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(3)

$Applying Kirchhoff's law$

$\frac{Q}{C} - L \frac{d I}{d t} = 0 \Rightarrow \frac{Q}{C} = L \frac{d I}{d t}$

$\Rightarrow Q = L C \frac{d}{d t} (- \frac{d Q}{d t}) = - L C \frac{d^{2} Q}{d t^{2}}$

Q 9 :

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values. [2013]

Q. If the direct transmission method with a cable of resistance $0.4 Ω {km}^{- 1}$ is used, the power dissipation (in %) during transmission is

20
30
40
50

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(1)

Step up transformer $\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{p}} \Rightarrow \frac{10}{1} = \frac{V_{s}}{4000}$

$∴$ $V_{s} = 40, 000 V$

Step down transformer $\frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}} = \frac{40, 000}{200} = \frac{200}{1}$

Q 10 :

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values. [2013]

Q. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is

200 : 1
150 : 1
100 : 1
50 : 1

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(2)

$Power P = V \times I$

$\Rightarrow I = \frac{P}{V} = \frac{600 \times 1000}{4000} = 150 A$

$Total resistance = 0.4 \times 20 = 8 Ω$

$∴$ $Power dissipated as heat = I^{2} R = {(150)}^{2} \times 8$

$= 180, 000 W = 180 kW$

$∴$ $% loss = \frac{180}{600} \times 100 = 30 %$

Topic Question Set

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Topic Question Set

Q 1 : In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 μC. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 μC. [IMAGE 1000] Q. The magnitude of q1 is ______. [2021]

Q 2 : In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 μC. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 μC. [IMAGE 1003] Q. The magnitude of q2 is _____. [2021]

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