NEET Physics Questions: Alternating Current

Pratcice NEET questions from all chapters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.

Physics Chemistry Botany Zoology

Please Login or Sign up to use advanced filters.

Question bank:

A transformer of efficiency 90% draws an input power of 4 kW. An electrical applience connected across the secondary draws a current of 6 A. The impedence of device is.........




No explanation available.

The core of a transformer is laminated so that.......





fact

In transformer, core is made of soft iron to reduce.....




No explanation available.

A primary winding of transformer has 500 turns whereas its secondary has 5000 turns. Primary is connected to ac supply of 20V, 50Hz The secondary output of....





The transformer works on the principle of mutual induction and the voltage transformation ratio is given by the formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Here, \(V_p\) is the primary voltage, \(V_s\) is the secondary voltage, \(N_p\) is the number of primary turns, and \(N_s\) is the number of secondary turns. Given \(V_p = 20V\), \(N_p = 500\), and \(N_s = 5000\), the secondary voltage \(V_s\) can be calculated as: \[ V_s = V_p \times \frac{N_s}{N_p} = 20V \times \frac{5000}{500} = 200V \] The frequency remains the same at 50 Hz. Hence, the correct answer is

200V, 50Hz

.

A step down transformer is connected to main supply 200 V to operate a 6V, 30 w bulb. The current in primary is.....





To find the current in the primary coil, we first need to determine the current in the secondary coil. The power consumed by the bulb is 30W and the voltage across it is 6V. Using the formula \( P = V \times I \), we get: \[ I_s = \frac{P}{V} = \frac{30W}{6V} = 5A \] The transformer is step-down, so the primary voltage is higher than the secondary. The current transformation ratio is given by \( \frac{I_s}{I_p} = \frac{V_p}{V_s} \). Given \(V_p = 200V\) and \(V_s = 6V\), we can calculate the primary current \(I_p\) as: \[ I_p = I_s \times \frac{V_s}{V_p} = 5A \times \frac{6V}{200V} = 0.15A \] Hence, the current in the primary is 0.15 A.
  • πŸŽ‰ Independence Day Offer: Get 25% off with code FREEDOM25! πŸŽ‰
  • πŸ“˜ Chapter progress tracker with auto color coding
  • πŸ” Weak‑area identification & detailed analytics
  • ✏️ Daily DPPs, mini & custom tests
  • ✨ Offline progress recorder & many more
Alternating current cannot be measured by dc ammeter because,




A DC ammeter measures the average value of current. For an alternating current (AC), the average value over a complete cycle is zero because AC alternates in direction and spends equal time in positive and negative half-cycles. Hence, a DC ammeter cannot measure AC.
The resistance of a coil for dc is in ohms. In ac, the resistance




In AC, the resistance of a coil is influenced by both its inherent resistance and reactance (inductive or capacitive). The reactance adds to the resistance, resulting in a higher effective resistance when compared to DC. This is why the resistance will increase in AC.

An alternating current of rms value 10 A is passed through a $ 12 \Omega $ resistance. The maximum potential difference across the resistor is,





To find the maximum potential difference across the resistor, we use the formula $V_{max} = I_{rms} imes R imes \sqrt{2}$. Given $I_{rms} = 10 ext{ A}$ and $R = 12 \\Omega$, we have $V_{max} = 10 imes 12 imes \sqrt{2} = 169.68 ext{ V}$. Therefore, the maximum potential difference is 169.68 V.
220 V,50 Hz, ac is applied to a resistor. The instantaneous value of voltage is




No explanation available.
The rmsvalue of anac of 50 Hz is 10 amp.The time takenbythe alternating current in reaching from zero to maximum value and the peak value of current will be,




For an AC current, the peak value $I_m$ can be calculated from the RMS value using $I_m = I_{rms} imes ext{√2} = 10 imes ext{√2} = 14.14$ A. The time taken to reach from zero to maximum value for an AC current of frequency 50 Hz is a quarter of the period, i.e., $T/4 = rac{1}{4f} = rac{1}{4 imes 50} = 5 imes 10^{-3}$ seconds.
If a current I given by $ I_0 sin ({wt - { \pi \over 2 } } )$ flows in an ac circuit across which an ac potential of $ E = E_0 sin \Omega t $ has been applied , then the power consumption p in the circuit will be




The power consumed in an AC circuit is given by $P = VI ext{cos}( heta)$ where $ΞΈ$ is the phase difference between the voltage and current. Here, $I = I_0 ext{sin}(Ο‰t - rac{Ο€}{2})$ and $E = E_0 ext{sin}(Ξ©t)$. Since the phase difference is $Ο€/2$, $ ext{cos}(Ο€/2) = 0$. Therefore, the power consumption $P = E_0 I_0 ext{cos}(Ο€/2) = 0$.
In general in an alternating current circuit.




No explanation available.
An alternating current is given by the eqn $ I =I_1 cos wt + I_2 sin wb $ . The rms current is given by




To find the RMS value of the given alternating current, we need to use the formula for the RMS value of a function composed of two orthogonal components. Here, the given current is $I = I_1 ext{cos}( ext{wt}) + I_2 ext{sin}( ext{wb})$. The correct formula for the RMS value is $I_{ ext{rms}} = rac{1}{ ext{sqrt 2}} imes ext{sqrt}(I_1^2 + I_2^2)$. Thus, the correct option is $\frac{1}{\sqrt{2}}(I_1^2 + I_2^2)^{\frac{1}{2}}$.
In an ac circuit, the current is given by $ I = 5 sin [ 100t - { \pi \over 2 } ] $ and the ac potential is V = 200 sin 100t. Then the power consumption is,




In an AC circuit, power consumption is given by P = V_rms * I_rms * cos(Ο•), where Ο• is the phase difference between the voltage and the current. Here, the current is $I = 5 ext{sin}(100t - \frac{\pi}{2})$ and the voltage is $V = 200 ext{sin}(100t)$. The phase difference Ο• is $\frac{\pi}{2}$. For a phase difference of $\frac{\pi}{2}$, cos(Ο•) = cos($\frac{\pi}{2}$) = 0. Therefore, the power consumption is $0$ watts.
In ac circuit with voltage V and current I, the power dissipaled is.




The power dissipated in an AC circuit depends on the phase difference between voltage (V) and current (I). When V and I are in phase, the power is maximized, and when they are out of phase, the power is reduced. This relationship is given by the formula: \( P = VI \cos(\phi) \), where \( \phi \) is the phase angle between V and I.