Electric Circuits Test 1

Result

Electric Circuits Test 1

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
2 / -0.33

A 5 μF condenser is connected in series with a coil having inductance of 50 mH. If a 50 V source operating at resonance frequency causes a circuit current of 10 mA. What is the Q factor of the coil.(upto 2 decimals)

A
0.01-0.03

B
0.014-0.031

C
0.017-0.036

D
0.019-0.032

Solution

Given that
L = 50 × 10^-3 H
C = 5 × 10^-6 F
Resonant frequency, \({\omega _0} = \frac{1}{{\sqrt {LC} }}\)
\({\omega _0} = \frac{1}{{\sqrt {50 \times {{10}^{ - 3}} \times 5 \times {{10}^{ - 6}}} }} = 2000rad/sec\;\;\)
Given that, current = 10 mA
At resonance, \({I_0} = \frac{V}{R}\)
\(\Rightarrow 10 \times {10^{ - 3}} = \frac{{50}}{R}\)
⇒ R = 5 KΩ
Quality factor, \(Q = \frac{{{\omega _0}L}}{R} = \frac{{2000 \times 50 \times {{10}^{ - 3}}}}{{5 \times {{10}^3}}} = 0.02\)
Question 2
2 / -0.33

In a RC type low pass fitter, the cut-off frequency is 1500 Hz, Assuming the capacitance to be 500 nF. The value of the resistance is – (in Ω)

A
211-213

B
2114-2131

C
2117-2136

D
2119-2132

Solution

we know that,
Cut – off frequency \({f_c} = \frac{1}{{2\pi RC}}\)
\(\begin{array}{l} \Rightarrow R = \frac{1}{{2\pi {f_c}C}}\\ \Rightarrow R = \frac{1}{{2\pi {f_c}C}}\\ \Rightarrow R = \frac{1}{{2\pi \times 1500 \times 500 \times {{10}^{ - 9}}}} \end{array}\)
⇒ R = 212.2 Ω
Question 3
2 / -0.33

At a particular instant, the R-phase voltage of a balanced 3-phase system is +60 V, and y phase voltage is -120 V. The voltage of B phase at that instant is ______ (in V)

A
60-60

B
604-601

C
607-606

D
609-602

Solution

Given 3-phase system is balanced, let the instantaneous voltage is
V_R (t) = V_m cosωt
V_Y(t) = V_m cos(ωt - 120°)
V_B (t) = V_m cos(ωt + 120°)
V_mcosωt = 60 V
V_m cos(ωt - 120°) = -120 V
⇒ V_m [cosωt cos120 + sinωt sin120] = -120
\(\begin{array}{l} \Rightarrow {V_m}\left[ {{\rm{cos\omega t}}\left( { - \frac{1}{2}} \right) + sin\omega t\left( {\frac{{\sqrt 3 }}{2}} \right)} \right] = - 120\\ \Rightarrow 60\left( { - \frac{1}{2}} \right) + \frac{{\sqrt 3 }}{2}{V_m}sin\omega t = - 120\\ \Rightarrow 60\left( { - \frac{1}{2}} \right) + \frac{{\sqrt 3 }}{2}{V_m}sin\omega t = - 120 \Rightarrow 60\left( { - \frac{1}{2}} \right) + \frac{{\sqrt 3 }}{2}{V_m}sin\omega t = - 120\end{array}\)
V_m cosωt = 60 V
⇒ tanωt = - √3 ⇒ ωt = -60°
\(\begin{array}{l}V_m^2{\sin ^2}\omega t + V_m^2{\cos ^2}\omega t = {\left( {60\sqrt 3 } \right)^2} + {\left( {60} \right)^2} \Rightarrow {V_m} = 120\;V\\{V_B} = {V_m}\cos \left( {\omega t + 120} \right) = 120\cos \left( { - 60 + 120} \right) = 120 \times \frac{1}{2} = 60\;V\end{array}\)