An electric motor operating on a 60 V dc supply draws a current of 10 A. If the efficiency of the motor is 50%, the resistance of its winding is 

(1) 3Ω

(2) 6Ω

(3) 15Ω

(4) 30Ω

The potential difference V and the current i flowing through an instrument in an ac circuit of frequency f are given by $V=5\cos \omega t$ volts and I = 2 sin ωt amperes (where ω = 2πf). The power dissipated in the instrument is 

(1) Zero

(2) 10 W

(3) 5 W

(4) 2.5 W

In an ac circuit, V and I are given by V = 100 sin (100 t) volts, $I=100\sin \left(100t+\frac{{\displaystyle \pi }}{{\displaystyle 3}}\right)mA$. The power dissipated in circuit is 

(1) 10⁴ watt

(2) 10 watt

(3) 2.5 watt

(4) 5 watt

The impedance of a coil, when DC supply is replaced by AC supply:
1. will remain the same
2. will increase
3. will decrease
4. will be zero

A generator produces a voltage that is given by V = 240 sin 120 t, where t is in seconds. The frequency and r.m.s. voltage are 

(1) 60 Hz and 240 V

(2) 19 Hz and 120 V

(3) 19 Hz and 170 V

(4) 754 Hz and 70 V

The peak value of an alternating e.m.f. E is given by $E=E_0\cos \omega t$ is 10 volts and its frequency is 50 Hz. At time $t=\frac{1}{600}sec$, the instantaneous e.m.f. is

1. 10 V

2. $5\sqrt{3}V$

3. 5 V

4. 1 V

If a current I given by $I_0\sin \left(\omega t-\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)$ 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 

(1) $P=\frac{E_0{I}_0}{\sqrt{2}}$

(2) $P=\sqrt{2}{E}_0{I}_0$

(3) $P=\frac{E_0{I}_0}{2}$

(4) P = 0

An alternating current is given by the equation $i=i_1\cos \omega t+i_2\sin \omega t$. The r.m.s. current is given by

(1) $\frac{1}{\sqrt{2}}(i_1+i_2)$

(2) $\frac{1}{\sqrt{2}}{(i_i+i_2)}^2$

(3) $\frac{1}{\sqrt{2}}{(i_1^2+i_2^2)}^{1/2}$

(4) $\frac{1}{2}{(i_1^2+i_2^2)}^{1/2}$

A resistance of \(20~ \mathrm{ohms}\) is connected to a source of an alternating potential, \(V=220sin(100 \pi t).\) The time taken by the current to change from its peak value to its r.m.s value will be: 

Voltage and current in an ac circuit are given by $V=5\sin \left(100\pi t-\frac{{\displaystyle \pi }}{{\displaystyle 6}}\right)$ and $I=4\sin \left(100\pi t+\frac{{\displaystyle \pi }}{{\displaystyle 6}}\right)$ 

(1) Voltage leads the current by 30°

(2) Current leads the voltage by 30°

(3) Current leads the voltage by 60°

(4) Voltage leads the current by 60°