Alternating Current Test

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Alternating Current Test - 46

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Question 1
1 / -0

In alternating current:

A
The direction of current is always positive

B
The direction of current is always negative

C
The direction of current changes constantly

D
None of these

Solution

A current that changes direction after equal intervals of time is called an alternating current.
Question 2
1 / -0

A sinusoidal voltage $$V=200sin314t$$ is applied to a $$10\Omega$$ resistor. Find the peak voltage.

A
$$200V$$

B
$$400V$$

C
$$600V$$

D
$$800V$$

Solution

$$Answer:-$$ A
General sinusoidal voltage variation is given by:
$$V=V_0sin(\omega t)$$
Here, in this equation $$V_0$$ is peak voltage.
So, on comparing both equation we get:
$$V_0=200V ,\omega=314$$
Question 3
1 / -0

A $$100\Omega$$ resistor is connected to a $$220V, 50Hz$$ AC supply. Find $$rms$$ value of current in the circuit :

A
$$1.10A$$

B
$$2.20A$$

C
$$3.30A$$

D
$$4.40A$$

Solution

$$Answer:-$$ B
The rated voltage in bulb is rms voltage so using:
$$V_{rms}=I_{rms}R $$
$$I_{rms}=\dfrac{220}{100}=2.20A$$
Question 4
1 / -0

A light bulb is rated at $$100W$$ for a $$220V$$ supply. Find the rms current through the bulb:

A
$$0.25A$$

B
$$0.45A$$

C
$$0.65A$$

D
$$0.85A$$

Solution

$$Answer:-$$ B
Power is given by:
$$P=\dfrac { { V }^{ 2 } }{ R } \\ 100=\dfrac { { (220 })^{ 2 } }{ R } \\ R=484\Omega $$

The rated voltage in bulb is rms voltage so using:
$$V_{rms}=I_{rms}R $$
$$I_{rms}=\dfrac{220}{484}=0.45A$$
Question 5
1 / -0

A condenser of $$250 \mu F$$ is connected in parallel to a coil of inductance 0.16 mH, while its effective resistance is $$20 \Omega$$. Determine the resonant frequency.

A
$$9\times 10^4 Hz$$

B
$$16\times 10^7 Hz$$

C
$$8\times 10^5 Hz$$

D
$$9\times 10^3 Hz$$

Solution

For the given case we have standard formula for resonant frequency as:
$$f\:=\dfrac{1}{2\pi}\:\sqrt{\dfrac{1}{LC}-\dfrac{R^2}{L^2}}$$
$$f\:=\dfrac{1}{2\pi}\:\sqrt{\dfrac{1}{0.16\times 10^{-3}\times 250\times 10^{-6}}-\dfrac{400}{({0.16\times10^{-3}})^{2}}}$$
On directly putting the given values in he equation we get:
$$\omega\:=\:7.98\times10^5$$
Question 6
1 / -0

In R-L-C series circuit, the potential differences across each element is $$20V$$. Now the value of the resistance alone is doubled, then P.D. across R, L and C respectively.

A
$$20V, 10V, 10V$$

B
$$20V, 20V, 20V$$

C
$$20V, 40V, 40V$$

D
$$10V, 20V, 20V$$

Solution

Circuit is at resonance ($$VL = VC$$)
∴ circuit is purely resistance
Resistance is doubled, current in the circuit is half the initial value
∴ New current $$I'=I/2$$
∴ $$VR = 20 V$$ (equal to applied voltage earlier)
$$VL = 10V$$
$$VC = 10V$$
Question 7
1 / -0

In A.C. circuit in which inductance and capacitance are joined in series, current is found to be maximum when the value of inductance is $$0.5H$$ and the value of capacitance is $$8\mu F$$. The angular frequency of applied alternating voltage will be

A
400 Hz

B
500 Hz

C
4000 Hz

D
5000 Hz

Solution

Given,
Inductance and capacitance are joined in series, and inductance $$L=0.5H$$,
And capacitance $$C= 8\mu F=8\times10^{-6}F$$.
We know that,
$$\omega=\frac{1}{\sqrt{LC}}$$
$$\omega=\dfrac{1}{\sqrt{8\times 10^{-6}\times 0.5}}=\dfrac{1}{2\times 10^{-3}}=0.5\times 10^3HZ=500HZ$$
Question 8
1 / -0

The reactance of coil when used in an A.C. power supply ( 220volts, 50 cycles/sec) is 50ohms. The inductance of the coil is nearly

A
0.16 H

B
0.22 H

C
2.2 H

D
1.6 H

Solution

Given : $$f = 50$$ cycles/sec
Reactance $$X_L = 50\Omega$$
Inductance of the coil $$L = \dfrac{X_L}{w} = \dfrac{50}{2\pi \ f } = \dfrac{50}{2\pi\times 50} = 0.16 \ H$$
Question 9
1 / -0

A series combination of resistor ($$R$$) and capacitor ($$C$$) is connected to an A.C. source of angular frequency '$$\omega$$'. Keeping the voltage same, if the frequency is changed to $$\omega/3$$, the current becomes half of the original current. Then the ratio of the capacitive reactance and resistance at the former frequency is:

A
$$\sqrt{0.6}$$

B
$$\sqrt{3}$$

C
$$\sqrt{2}$$

D
$$\sqrt{6}$$

Solution

Let r.m.s. value of supply voltage be $$v$$.

Impedance of RC circuit is given by:
$$Z = \sqrt {R^2 + X_C^2} = \sqrt {R^2 + \dfrac{1}{\omega^2 C^2}}$$

Then, r.m.s. current in the circuit is:
$$ \displaystyle i = \cfrac{v}{Z} = \cfrac{ v } { \sqrt { R^2 + \frac {1}{\omega^2C^2} }}$$

$$\implies \cfrac{i_2}{i_1} = \cfrac{\sqrt {R^2 + \frac{1}{\omega^2 C^2}} } { \sqrt {R^2 + \frac{1}{(\omega/3)^2 C^2}} }$$
Given : $$\dfrac{i_2}{i_1} = \dfrac{1}{2}$$ and $$\omega_2 = \omega/3$$
On substituting values and simplifying, we get:
$$ \cfrac{1}{4} = \cfrac { 1+\omega^2 R^2 C^2 }{9 + \omega^2 R^2 C^2}$$
$$\implies \omega RC = \sqrt{5/3}$$

Ratio at capacitive reactance and resistance in original frequency is:
$$\cfrac{X_C}{R} = \cfrac{1}{\omega RC} = \sqrt {3/5} = \sqrt {0.6}$$
Question 10
1 / -0

In a series LCR circuit, resonance occurring at 105Hz. At that time, the potential difference across the 100 resistance is 40V while the potential difference across the pure inductor is 30v. The inductance L of the inductor is equal to

A
$$2.0 \times 10^{-4}$$

B
$$3.0 \times 10^{-4}$$

C
$$1.2 \times 10^{-4}$$

D
$$2.4 \times 10^{-4}$$

Solution

In a series LCR circuit, resonance occurring at 105Hz. At that time, the potential difference across the 100 resistance is 40V while the potential difference across the pure inductor is 30v. The inductance L of the inductor is equal to
$$Current(i)=\dfrac{V_r}{R}=\dfrac{100}{40}=2.5A$$
the inductor value is $$V_L=i*(wL)$$
$$L=\dfrac{30}{2.5*2*\pi*105}=1.2*10^{-4}H$$

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Alternating Current Test - 46

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Alternating Current Test - 46

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Question 1

The direction of current is always positive

The direction of current is always negative

The direction of current changes constantly

None of these

Solution

Question 2

$$200V$$

$$400V$$

$$600V$$

$$800V$$

Solution

Question 3

$$1.10A$$

$$2.20A$$

$$3.30A$$

$$4.40A$$

Solution

Question 4

$$0.25A$$

$$0.45A$$

$$0.65A$$

$$0.85A$$

Solution

Question 5

$$9\times 10^4 Hz$$

$$16\times 10^7 Hz$$

$$8\times 10^5 Hz$$

$$9\times 10^3 Hz$$

Solution

Question 6

$$20V, 10V, 10V$$

$$20V, 20V, 20V$$

$$20V, 40V, 40V$$

$$10V, 20V, 20V$$

Solution

Question 7

400 Hz

500 Hz

4000 Hz

5000 Hz

Solution

Question 8

0.16 H

0.22 H

2.2 H

1.6 H

Solution

Question 9

$$\sqrt{0.6}$$

$$\sqrt{3}$$

$$\sqrt{2}$$

$$\sqrt{6}$$

Solution

Question 10

$$2.0 \times 10^{-4}$$

$$3.0 \times 10^{-4}$$

$$1.2 \times 10^{-4}$$

$$2.4 \times 10^{-4}$$

Solution

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