A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
300 V, 15 A
450 V, 15 A
450 V, 13.5 A
450 V, 13.5 A
In an AC circuit, an alternating voltage 100t volt is connected to a capacitor of capacity 1μF. The rms value of the current in the circuit is
100 mA
200 mA
20 mA
20 mA
An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is
π/4
π/2
zero
zero
In the given figure, a diode D is connected to an external resistance R = 100 Ω and an e. m. f of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be
30 mA
40 mA
20 mA
20 mA
A 220 V input is supplied to a transformer. The output circuit draws a current 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
3.6 A
2.8 A
2.5 A
2.5 A
The two ends f a rod of length L and a uniform cross -sectional area A are kept at two temperatures T1 and T2 (T2 > T1) .The rate of heat transfer, dQ/dt, through the rod in steady state is given by
A galvanometer having a coil resistance of 60 Ω shows full-scale deflection when a current of 1.0 A passes through it. It can be converted into an ammeter to read currents up to 5.0 A by
putting in parallel a resistance of 240 Ω
putting in series a resistance of 15 Ω
putting in series a resistance of 240 Ω
putting in series a resistance of 240 Ω
In an AC circuit, the emf (e) and the current (i) at any instant are given respectively by
e = E0 sin ωt
i = Io sin (ωt -Φ)
The average power in the circuit over one cycle of AC is
In the circuit shown, the current through the 4 Ω resistor is 1 A when the points P and M are connected to a DC voltage source. The potential difference between the point M and N is
1.5 V
1.0 V
0.5 V
0.5 V
D.
0.5 V
In parallel resistances, the potential difference across them is same.
Potential difference across PM
V = 4 x 1 = 4 Volt (ie, across 4 Ω)
Equivalent resistance of lower side arm,
Now the circuit can be shown as
Let current I flow in lower branch, so
1.25 I = 4V
I = 4/1.25 = 3.2 A
Therefore, 3.2 A current flows in 1 Ω resistance, hence potential difference between M and N is
V' = 3.2 x 1 = 3.2 volt