The network shown in the figure is a part of a complete circuit. If at a certain instant the current i is 5 A and is decreasing at the rate of 10³A/s then \(V_{B}-V_{A}\) is :
An ideal coil of 10H is connected in series with a resistance of 5Ω and a battery of 5V. 2 seconds after the connection is made, the current flowing in amperes in the circuit is :
In step-down transformer of turn ratio 6 : 1, an alkali accumulator of 12 V; 3A is connected to the primary. Then the voltage developed in its secondary is :
The voltage developed in the secondary is zero.
A transformer operates on the principle of mutual induction, which requires a changing magnetic flux produced by Alternating Current (AC). Since an alkali accumulator provides Direct Current (DC), the magnetic flux remains constant, and no electromotive force (EMF) is induced in the secondary coil.
The expression for magnetic induction inside a solenoid of length \(L\) carrying a current \(I\) and having \(N\) number of turns is:
Magnetic field inside a long solenoid is given by \(B = \mu_0 n I\), where \(n = \frac{N}{L}\) is the number of turns per unit length. Therefore, \(B = \mu_0 \frac{N}{L} I\).
A coil has 200 turns and area of \(70\text{ cm}^2\). The magnetic field perpendicular to the plane of the coil is \(0.3\text{ Wb m}^{-2}\) and takes \(0.1\text{ s}\) to rotate through \(180^\circ\). The value of the induced E.M.F. will be:
Change in flux \(\Delta \Phi = 2NBA = 2 \times 200 \times 0.3 \times (70 \times 10^{-4}) = 0.84\text{ Wb}\). Induced EMF is \(e = \frac{\Delta \Phi}{\Delta t} = \frac{0.84}{0.1} = 8.4\text{ V}\).
If a current is passed through a spring then the spring will:
Current flows in the same direction in adjacent turns of the spring. Since parallel currents in the same direction attract each other, the turns of the spring are pulled closer, causing the spring to compress.
In a step up transformer, the voltage in the primary is \(220\text{ V}\) and the current is \(5\text{ A}\). The secondary voltage is found to be \(22000\text{ V}\). The current in the secondary coil (neglect losses) is:
In an ideal transformer, input power equals output power: \(V_p I_p = V_s I_s\). Therefore, \(I_s = \frac{V_p I_p}{V_s} = \frac{220 \times 5}{22000} = 0.05\text{ A}\).
An aeroplane in which the distance between the tips of the wings is \(50\text{ m}\) is flying horizontally with a speed of \(360\text{ km/hr}\) over a place where the vertical component of earth’s magnetic field is \(2 \times 10^{-4}\text{ Wbm}^{-2}\). The potential difference between the tips of the wings would be:
Induced EMF is \(e = B_v l v\). Converting speed: \(v = 360\text{ km/h} = 100\text{ m/s}\). Thus, \(e = (2 \times 10^{-4}) \times 50 \times 100 = 1.0\text{ V}\).
A current of \(2\text{ A}\) is increasing at a rate of \(4\text{ A/s}\) through a coil of inductance \(2\text{ H}\). The energy stored in the inductor per unit time in given instant is:
Formula for rate of change of energy in an inductor is \(\frac{dU}{dt} = LI\frac{dI}{dt}\). Given \(L = 2\text{ H}\), \(I = 2\text{ A}\), and \(\frac{dI}{dt} = 4\text{ A/s}\), we get \(\frac{dU}{dt} = 2 \times 2 \times 4 = 16\text{ J/s}\).
Two coils of self inductance \(2\text{ mH}\) and \(8\text{ mH}\) are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:
For complete coupling, the coupling coefficient \(k = 1\). The mutual inductance is given by \(M = k\sqrt{L_1 L_2} = \sqrt{2 \times 8} = 4\text{ mH}\).