Alternating current practice · Mixed difficulty

Option D is correct. The time to reach the peak value is one-quarter of the period, calculated as $\frac{1}{4f} = \frac{1}{240}$ s.

Option C (0.6, 4.8 W) is correct because the power factor is calculated as R/Z and the power dissipated is calculated as V²*R/Z².

Option C is correct. In a purely resistive AC circuit, the current and voltage are always in phase.

Option C (14.14A) is correct because the peak value is calculated by multiplying the RMS value by $\sqrt{2}$.

Option A (240 V,5 A) is correct because the voltage is doubled and the current is halved due to the turns ratio of the transformer.

Option C is correct because at resonance in a series L-C-R circuit, the inductive reactance equals the capacitive reactance, resulting in a phase difference of zero between current and voltage.

Option A ($\frac{{20}}{{\sqrt {13} }}$) is correct because the current is calculated using the impedance at 100 Hz.

Option B (54μF) is correct because the capacitance is calculated to cancel out the inductive reactance and achieve a unity power factor.

Option C (π/4) is correct because the phase difference is given by the arctangent of the ratio of inductive reactance to resistance, which are equal in this case.

Option C (Zero) is correct because an ideal capacitor does not dissipate energy; it only stores energy.

Option B (Greater than one) is correct because a step-up transformer increases the voltage, which means the transformation ratio (secondary voltage/primary voltage) is greater than one.

Option A (1 W) is correct because the maximum rate at which energy is stored in the magnetic field is equal to 1/2 * L * I², where I is the maximum current.

Option C is correct. A choke coil with negligible resistance dissipates no power because the voltage and current are 90 degrees out of phase.

Option C (${v_ \circ } = v = \frac{{50}}{\pi }Hz$) is correct because both the resonant frequency and the frequency of the AC source are calculated to be $\frac{{50}}{\pi }Hz$.

Option D is correct. The average power dissipation in an AC circuit is given by $\frac{1}{2} E_0 I_0 \cos \phi$, where $\phi$ is the phase difference between voltage and current.

Option C ($2.82 {~A}$) is correct because an AC ammeter reads the RMS value of the AC component, and the total current is the RMS value of the combined DC and AC currents.

Option A ($45\,V$) is correct because the voltage in the secondary coil is calculated using the transformation ratio of the transformer.

Option A (1.25 watt) is correct because the power dissipated in an AC circuit is given by $P = V_{rms} I_{rms} \cos{\phi}$, where $\phi$ is the phase difference between voltage and current.

Option D ($\sqrt {1/LC}$) is correct because the current in a series LCR circuit is maximum at resonance, which occurs when the angular frequency equals $\sqrt {1/LC}$.

Option C (Either purely inductive or purely capacitive) is correct because in purely inductive or capacitive circuits, the voltage and current are out of phase by 90 degrees, resulting in zero average power dissipation.

Option B is correct. As the speed of a DC motor increases, the back EMF also increases, reducing the armature current.

Option A (${E^2}R/ {{R^2} + {{\left( {L\omega - \frac{{1}}{{C\omega }}} \right)}^2}} )$ is correct because it represents the power dissipated in an LCR series circuit, derived from the formula $P = \frac{{E^2}}{Z} \cos \phi$, where Z is the impedance and $\cos \phi$ is the power factor.

Option C (${10^{ - 3}}s$) is correct because the time taken for the change of current is calculated using the formula $E = -M \frac{dI}{dt}$, resulting in ${10^{ - 3}}s$.

Option C (if voltage applied is $V=V_m \sin \omega t$ then the circuit must be containing resistance and capacitance) is correct because the current has both sine and cosine components, indicating a phase shift that can only be achieved with both resistance and capacitance.

Option C ($60\;volts,\;200\;Hz$) is correct because the voltage equation can be simplified using trigonometric identities to find the maximum voltage and frequency.

Option B (mutual induction) is correct because a transformer works by transferring energy from the primary to the secondary coil through mutual induction.

Option B (500 W) is correct because the power dissipation is calculated as P = Vrms * Irms, where Vrms and Irms are the root mean square values of voltage and current.

Option B (Zero) is correct because a pure capacitor does not dissipate power; it only stores energy.

Option A (50 V) is correct because the supply voltage is the vector sum of the voltage across the resistor and the net voltage across the inductor and capacitor.

Option B (5 mH) is correct because the self-inductance is calculated using the formula $E = -L \frac{dI}{dt}$, resulting in 5 mH.