1
JEE Advanced 2023 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
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A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5 ~\mathrm{rad}~ \mathrm{s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega=8 \times 10^4 ~\mathrm{rad} ~\mathrm{s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L=50 ~\mathrm{mH}$, match each entry in List-I with an appropriate value from List-II and choose the correct option.

List - I List - II
(P) $I_0$ in $\mathrm{mA}$ (1) 44.4
(Q) The quality factor of the circuit (2) 18
(R) The bandwidth of the circuit in $\mathrm{rad}~ \mathrm{s}^{-1}$ (3) 400
(S) The peak power dissipated at resonance in Watt (4) 2250
(5) 500
A
$P \rightarrow 2, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 1$
B
$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2$
C
$P \rightarrow 4, Q \rightarrow 5, R \rightarrow 3, S \rightarrow 1$
D
$P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 5$
2
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0
Consider a simple RC circuit as shown in Figure 1.

Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e. charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC.

Process 2 : In a different process the voltage is first set to $${{{V_0}} \over 3}$$ and maintained for a charging time T >> RC. Then, the voltage is raised to $${{2{V_0}} \over 3}$$ without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1.

These two processes are depicted in Figure 2.
JEE Advanced 2017 Paper 2 Offline Physics - Alternating Current Question 10 English Comprehension
In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by
A
EC = ED ln2
B
EC = ED
C
EC = 2ED
D
EC = $${1 \over 2}$$ED
3
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0
Consider a simple RC circuit as shown in Figure 1.

Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e. charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC.

Process 2 : In a different process the voltage is first set to $${{{V_0}} \over 3}$$ and maintained for a charging time T >> RC. Then, the voltage is raised to $${{2{V_0}} \over 3}$$ without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1.

These two processes are depicted in Figure 2.

JEE Advanced 2017 Paper 2 Offline Physics - Alternating Current Question 9 English Comprehension
In Process 2, total energy dissipated across the resistance ED is
A
$${E_D} = {1 \over 3}\left( {{1 \over 2}CV_0^2} \right)$$
B
$${E_D} = 3\left( {{1 \over 2}CV_0^2} \right)$$
C
$${E_D} = 3CV_0^2$$
D
$${E_D} = {1 \over 2}CV_0^2$$
4
JEE Advanced 2013 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

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.

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 of 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

A
200 : 1
B
150 : 1
C
100 : 1
D
50 : 1
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