RMS Value

Q. Why rms value of voltages for square wave become root two times of sine wave proove it

Q. 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. $\frac{1}{\sqrt{2}}(i_1+i_2{)}^2$
3. $\frac{1}{\sqrt{2}}(i_1^2+i_2^2{)}^{\frac{1}{2}}$
4. $\frac{1}{2}(i_1^2+i_2^2{)}^{\frac{1}{2}}$

Q. The root mean square value of the alternating current is equal to

1. Twice the peak value
2. Half the peak value
3. $\frac{1}{\sqrt{2}}$ times the peak value
4. Equal to the peak value

Q. In a circuit, the value of the alternating current is measured by hot wire ammeter as 10 ampere. Its peak value will be

1. 10 A
2. 20 A
3. 14.14 A
4. 7.07 A

Q. In an ac circuit, the r.m.s. value of current, $I_{rms}$ is related to the peak current, $I_0$ by the relation

1. $I_{rms}=\frac{1}{\pi }{I}_0$
2. $I_{rms}=\frac{1}{\sqrt{2}}{I}_0$
3. $I_{rms}=\sqrt{2}{I}_0$
4. $I_{rms}=\pi I_0$

Q. The maximum value of a.c. voltage in a circuit is 707V. Its rms value is

1. 70.7 V
2. 100 V
3. 500 V
4. 707 V

Q. RMS value of an alternating voltage is $\frac{1}{\sqrt{2}}$ times the value of peak voltage and is the value of the equivalent DC voltage.

1. False
2. True

Q. The r.m.s. value of an ac of 50 Hz is 10 amp. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be

1. $2\times {10}^{-2}$ sec and 14.14 amp
2. $1\times {10}^{-2}$ sec and 7.07 amp
3. $5\times {10}^{-3}$ sec and 7.07 amp
4. $5\times {10}^{-3}$ sec and 14.14 amp

Q. A 280 ohm electric bulb is connected to 200V electric line. The peak value of current in the bulb will be

1. About one ampere
2. Zero
3. About two ampere
4. About four ampere

Q. The rms value of an alternating current, which when passed through a resistor produces heat three times of that produced by a direct current of 2A in the same resistor, is

1. 6A
2. 3A
3. 2A
4. 2$\sqrt{3}$A

Q. A $10\Omega$ resistor, 2H inductor and 2.5 mF capacitor are all connected in parallel across a 10v, $10rad{s}^{-1}$ source. Find the rms current through the source.

1. 0.5 A
2. 2 A
3. 1.03 A
4. 0.2 A

Q. In a series LCR circuit $R=100\Omega$. When only c is removed the current leads voltage by ${60}^{\circ }$. When only L is removed current logs voltage by ${45}^{\circ }$. Find the rms current in the LCR circuit.

1. 5 A
2. 0.2 A
3. 1.6 A
4. 4 A

Q.

In a certain circuit current changes with time according to $i=2\sqrt{t}.$ r.m.s. value of current between t=2 to t=4s will be

1. $3A$

2. $3\sqrt{3}A$

3. $2\sqrt{3}A$

4. $(2-\sqrt{2})A$

Q. The frequency of an alternating voltage is 50 cycles/sec and its amplitude is 120V. Then the r.m.s. value of voltage is

1. 101.3V
2. 84.8V
3. 70.7V
4. 56.5V

Q. The current decaying in a circuit has a function $i=I_0{e}^{-t/\tau }$ where $\tau$ is the time constant of the circuit. Find $rms$ value of the current for the period $t=0\text{to}t=\tau$.

Q. Match the following

$\begin{array}{cc}Currents& r.ms.values\\ \left(1\right)x_{0}sin\omega t& \left(i\right)x_0\\ \left(2\right)x_{0}sin\omega tcos\omega t& \left(ii\right)\frac{x_0}{\sqrt{2}}\\ \left(3\right)x_{0}sin\omega t+x_{0}cos\omega t& \left(iii\right)\frac{x_0}{\left(2\sqrt{2}\right)}\end{array}$

1. 1. (i), 2. (ii), 3(i)
2. 1. (ii), 2. (i), 3. (ii)
3. 1. (ii), 2. (iii), 3. (i)
4. 1. (ii), 2. (ii), 3. (i)

Q.

In a certain circuit current changes with time according to $i=2\sqrt{t}.$ r.m.s. value of current between t=2 to t=4s will be

1. $3A$

2. $3\sqrt{3}A$

3. $2\sqrt{3}A$

4. $(2-\sqrt{2})A$

Q. A 220 V main supply is connected to a resistance of 100k $\Omega$ . The rms current is

1. 22 mA
2. 2.2mA
3. 220mA
4. 10 mA