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A coil of inductance 8.4 mH and resistance $ 6 \Omega $ is connected to a 12 V battery. The current in the coil is 1A in the time..........
$ I_0 = 12/6 = 2A . $ Current becomes half in time t = 0.693 L/R = 1ms
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
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 $θ$ is the phase difference between the voltage and current. Here, $I = I_0 ext{sin}(ωt - rac{π}{2})$ and $E = E_0 ext{sin}(Ωt)$. Since the phase difference is $π/2$, $ ext{cos}(π/2) = 0$. Therefore, the power consumption $P = E_0 I_0 ext{cos}(π/2) = 0$.
In general in an alternating current circuit.
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(Ï•), where Ï• 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 Ï• is $\frac{\pi}{2}$. For a phase difference of $\frac{\pi}{2}$, cos(Ï•) = cos($\frac{\pi}{2}$) = 0. Therefore, the power consumption is $0$ watts.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.