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Alternating Current

NEET Physics · 95 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

Q1
An ac circuit contains a resistance of 1kΩ1 \mathrm{k} \Omega, a capacitor of 0.1μ F0.1 \mu \mathrm{~F} and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately:
A 13.5 kHz
B 10.1 kHz
C 20.7 kHz
D 15.9 kHz
Correct Answer
Option D
Solution

Resonance frequency

f0=12πLC=12π1×10−7×10−3=15.9kHz\begin{aligned} & f_0=\frac{1}{2 \pi \sqrt{L C}} \\ & =\frac{1}{2 \pi \sqrt{1 \times 10^{-7} \times 10^{-3}}} \\ & =15.9 \mathrm{kHz} \end{aligned}
Q2
The peak value of an alternating current is 5 A and frequency is 60 Hz . How long will the current, starting from zero, take to reach the peak value?
A 1120 s\dfrac{1}{120} \mathrm{~s}
B 160 s\dfrac{1}{60} \mathrm{~s}
C 130 s\dfrac{1}{30} \mathrm{~s}
D 1240 s\dfrac{1}{240} \mathrm{~s}
Correct Answer
Option D
Solution

Alternating current,

i=ipeak sin⁡(ωt)i=i_{\text{peak }} \sin (\omega t)

where ω=2πf=2π×60=120πrad/s\omega=2 \pi f=2 \pi \times 60=120 \pi \mathrm{rad} / \mathrm{s}

⇒i=5sin⁡(120πt)⇒5=5sin⁡(120πt)⇒sin⁡(120πt)=1=sin⁡(π2)∴120πt=π2⇒t=1240 s\begin{aligned} & \Rightarrow \quad i=5 \sin (120 \pi t) \\ & \Rightarrow \quad 5=5 \sin (120 \pi t) \\ & \Rightarrow \quad \sin (120 \pi t)=1=\sin \left(\frac{\pi}{2}\right) \\ & \therefore \quad 120 \pi t=\frac{\pi}{2} \\ & \Rightarrow \quad t=\frac{1}{240} \mathrm{~s} \end{aligned}
Q3
To an ac power supply of 220 V at 50 Hz , a resistor of 20Ω20 \Omega, a capacitor of reactance 25Ω25 \Omega and an inductor of reactance 45Ω45 \Omega are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively
A 15.6 A and 30∘30^{\circ}
B 15.6 A and 45∘45^{\circ}
C 7.8 A and 30∘30^{\circ}
D 7.8 A and 45∘45^{\circ}
Correct Answer
Option D
Solution

To determine the current in the circuit and the phase angle between the current and voltage, follow these steps: Components and their Reactance: Inductive reactance, XL=45 Ω X_L = 45 \, \Omega Capacitive reactance, XC=25 Ω X_C = 25 \, \Omega Resistance, R=20 Ω R = 20 \, \Omega Calculate Current (I): The current through the series circuit can be calculated using the formula: I=V(XL−XC)2+R2 I = \dfrac{V}{\sqrt{(X_L - X_C)^2 + R^2}} Substituting the given values: I=220(45−25)2+202=220202+202 I = \dfrac{220}{\sqrt{(45 - 25)^2 + 20^2}} = \dfrac{220}{\sqrt{20^2 + 20^2}} Simplifying further: I=220400+400=220800=2202200=2202100×2=220202=112 I = \dfrac{220}{\sqrt{400 + 400}} = \dfrac{220}{\sqrt{800}} = \dfrac{220}{2\sqrt{200}} = \dfrac{220}{2\sqrt{100 \times 2}} = \dfrac{220}{20\sqrt{2}} = \dfrac{11}{\sqrt{2}} Therefore, the current is approximately: I=7.779 A (≈7.8 A) I = 7.779 \, \text{A} \, (\approx 7.8 \, \text{A}) Calculate Phase Angle (ϕ\phi): The phase angle can be calculated using the tangent of the angle: tan⁡ϕ=XL−XCR=45−2520=2020=1 \tan \phi = \dfrac{X_L - X_C}{R} = \dfrac{45 - 25}{20} = \dfrac{20}{20} = 1 Hence, the phase angle ϕ\phi is: ϕ=45∘ \phi = 45^{\circ} Thus, the current in the circuit is approximately 7.8 A, and the phase angle is 45∘45^{\circ}.

Q4
In the circuit shown below, the inductance LL is connected to an ac source. The current flowing in the circuit is I=I0sin⁡ωtI=I_0 \sin \omega t. The voltage drop (VL)\left(V_L\right) across LL is
A ωLI0sin⁡ωt\omega L I_0 \sin \omega t
B I0ωLsin⁡ωt\dfrac{I_0}{\omega L} \sin \omega t
C I0ωLcos⁡ωt\dfrac{I_0}{\omega L} \cos \omega t
D ωLI0cos⁡ωt\omega L I_0 \cos \omega t
Correct Answer
Option D
Solution
VLV_L

leads current

II

by

π2\frac{\pi}{2}
∴VL=V0sin⁡(ωt+π2)(∵I=I0sin⁡ωt)V0=I0XL⇒VL=I0XLcos⁡(ωt)=I0ωLcos⁡(ωt)\begin{aligned} & \therefore V_L=V_0 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left(\because I=I_0 \sin \omega t\right) \\ & V_0=I_0 X_L \\ & \Rightarrow V_L=I_0 X_L \cos (\omega t)=I_0 \omega L \cos (\omega t) \end{aligned}
Q5
A step up transformer is connected to an ac mains supply of 220 V220 \mathrm{~V} to operate at 11000 V,8811000 \mathrm{~V}, 88 watt. The current in the secondary circuit, ignoring the power loss in the transformer, is
A 8 mA
B 4 mA
C 0.4 A
D 4 A
Correct Answer
Option A
Solution

Let's analyze the problem step by step.

We are given a step-up transformer with the following specifications: Primary Voltage (V p ) =

220 V220 \mathrm{~V}

Secondary Voltage (V s ) =

11000 V11000 \mathrm{~V}

Power (P) =

8888

watt The power input to the transformer is equal to the power output, assuming there is no loss in the transformer.

Hence,

Pprimary=Psecondary=P=88 WP_{primary} = P_{secondary} = P = 88 \mathrm{~W}

The power equations can be written as:

Pprimary=VpIpP_{primary} = V_p I_p
Psecondary=VsIsP_{secondary} = V_s I_s

Given the power in the secondary winding is

88 W88 \mathrm{~W}

and the secondary voltage is

11000 V11000 \mathrm{~V}

, we can find the secondary current

IsI_s

using the formula:

Psecondary=VsIsP_{secondary} = V_s I_s

Rearranging to solve for

IsI_s

gives:

Is=PVsI_s = \frac{P}{V_s}

Substituting the given values:

Is=88 W11000 VI_s = \frac{88 \mathrm{~W}}{11000 \mathrm{~V}}

Simplifying this expression:

Is=8811000 AI_s = \frac{88}{11000} \mathrm{~A}
Is=0.008 AI_s = 0.008 \mathrm{~A}
Is=8 mAI_s = 8 \mathrm{~mA}

Therefore, the current in the secondary circuit, ignoring the power loss in the transformer, is 8 mA .

The correct option is Option A: 8 mA .

Q6
The amplitude of the charge oscillating in a circuit decreases exponentially as Q=Q0e−Rt/2LQ=Q_0 e^{-R t/2 L}, where Q0Q_0 is the charge at t=0 st=0 \mathrm{~s}. The time at which charge amplitude decreases to 0.50Q00.50 Q_0 is nearly: [Given that R=1.5Ω,L=12 mH,ln⁡(2)=0.693R=1.5 \Omega, L=12 \mathrm{~mH}, \ln (2)=0.693]
A 19.01 ms
B 11.09 ms
C 19.01 s
D 11.09 s
Correct Answer
Option B
Solution

The given equation for the amplitude of the charge oscillating in a circuit is:

Q=Q0e−Rt2LQ = Q_0 e^{- \frac{R t}{2 L} }

We need to find the time,

tt

, at which charge amplitude decreases to

0.50Q00.50 Q_0

. So, we set

Q=0.50Q0Q = 0.50 Q_0

:

0.50Q0=Q0e−Rt2L0.50 Q_0 = Q_0 e^{- \frac{R t}{2 L} }

Divide both sides by

Q0Q_0

:

0.50=e−Rt2L0.50 = e^{- \frac{R t}{2 L} }

Take the natural logarithm on both sides to solve for

tt

:

ln⁡(0.50)=−Rt2L\ln(0.50) = - \frac{R t}{2 L}

Recall that

ln⁡(0.50)=−ln⁡(2)\ln(0.50) = - \ln(2)

. Substituting the given value

ln⁡(2)=0.693\ln(2) = 0.693

, we get:

−0.693=−Rt2L-0.693 = - \frac{R t}{2 L}

Remove the negative signs from both sides:

0.693=Rt2L0.693 = \frac{R t}{2 L}

Now, solve for

tt

:

t=2L⋅0.693Rt = \frac{2 L \cdot 0.693}{R}

Substituting the given values

R=1.5ΩR = 1.5 \Omega

and

L=12 mH=12×10−3 HL = 12 \mathrm{~mH} = 12 \times 10^{-3} \mathrm{~H}

, we have:

t=2×12×10−3⋅0.6931.5t = \frac{2 \times 12 \times 10^{-3} \cdot 0.693}{1.5}

Calculate the numerator and denominator:

t=16.632×10−31.5t = \frac{16.632 \times 10^{-3}}{1.5}

Finally, compute the value of

tt

:

t=11.09×10−3 s=11.09 mst = 11.09 \times 10^{-3} \mathrm{~s} = 11.09 \mathrm{~ms}

Therefore, the time at which the charge amplitude decreases to

0.50Q00.50 Q_0

is nearly 11.09 ms. Thus, the correct answer is: Option B: 11.09 ms

Q7
In an ideal transformer, the turns ratio is NPNS=12\dfrac{N_P}{N_S}=\dfrac{1}{2}. The ratio VS:VPV_S: V_P is equal to (the symbols carry their usual meaning) :
A 1:21: 2
B 2:12: 1
C 1:11: 1
D 1:41: 4
Correct Answer
Option B
Solution

An ideal transformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils.

The primary coil (with

NPN_P

turns) is connected to the input voltage (

VPV_P

), and the secondary coil (with

NSN_S

turns) delivers the output voltage (

VSV_S

).

The relationship between the input (primary) voltage and output (secondary) voltage in a transformer is determined by the ratio of the number of turns in the primary coil to the number of turns in the secondary coil, expressed by the formula:

VSVP=NSNP\frac{V_S}{V_P} = \frac{N_S}{N_P}

In this case, the given turns ratio is

NPNS=12\frac{N_P}{N_S} = \frac{1}{2}

. To find the correct expression for

VSVP\frac{V_S}{V_P}

, we need to invert the given ratio to reflect the relationship with respect to the voltages. This means:

NSNP=2\frac{N_S}{N_P} = 2

Therefore, substituting into the voltage ratio equation:

VSVP=2\frac{V_S}{V_P} = 2

This indicates that the secondary voltage (

VSV_S

) is twice the primary voltage (

VPV_P

). If we were to express the ratio

VS:VPV_S:V_P

based on this, it would be: Option B

2:12: 1

This is the correct answer as it represents that, in accordance with the transformer's turns ratio, the voltage in the secondary coil is twice that in the primary coil.

Q8
A 10μF10 \mu \mathrm{F} capacitor is connected to a 210 V,50 Hz210 \mathrm{~V}, 50 \mathrm{~Hz} source as shown in figure. The peak current in the circuit is nearly (π=3.14)(\pi=3.14) :
A 0.58 A
B 0.93 A
C 1.20 A
D 0.35 A
Correct Answer
Option B
Solution

Capacitive Reactance

XC=1ωC=12πfC=12×3.14×50×10×10−6=10003.14\begin{aligned} X_C =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\ =\frac{1000}{3.14}\end{aligned}
Vrms=210 Virms=VrmsXC=210XC Peak current =2ims=2×2101000×3.14=0.932=0.93 A\begin{aligned} & V_{\mathrm{rms}}=210 \mathrm{~V} \\ & i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C} \\ & \text{ Peak current }=\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\ & \qquad =0.93 \mathrm{~A} \end{aligned}
Q9
The maximum power is dissipated for an ac in a/an:
A resistive circuit
B LC circuit
C inductive circuit
D capacitive circuit
Correct Answer
Option A
Solution

Power dissipated is maximum of purely resistive circuit.

Q10
For very high frequencies, the effective impedance of the circuit (shown in the figure) will be:-
A 4 Ω\Omega
B 6 Ω\Omega
C 1 Ω\Omega
D 3 Ω\Omega
Correct Answer
Option D
Solution

As frequency is very high

XC≈0\mathrm{X_C \approx 0}
XL→α\mathrm{X_L\to\alpha}

Effective circuit will be Effective impedance of circuit will be = 3

Ω\Omega
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