Network Elements and the Concept of Circuit
- D.C. component of the waveform shown below is—
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D. C. component = 1 
f(t) dt T T = 1 
π 2π 0.dt 
2π π = 10 – cos t π + 0 2π 0 = 10 – cos 2π π + cos 0 2π 2π = 10 
1 + 1 
2π = 10 π = 10 π . 2π 2π = 10 = 3.18 amp. π Correct Option: C
D. C. component = 1 f(t) dt T T = 1 π 2π 0.dt 2π π = 10 – cos t π + 0 2π 0 = 10 – cos 2π π + cos 0 2π 2π = 10 1 + 1 2π = 10 π = 10 π . 2π 2π = 10 = 3.18 amp. π
- D.C. component of the waveform shown below is—
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NA
Correct Option: B
NA
- If 120 C of charge passes through an electric conduction in 30 sec, the current in the conductor is—
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i = dq = 120 = 4 A dt 30 Correct Option: D
i = dq = 120 = 4 A dt 30
- The energy required to move 50 coulomb through 3 V is—
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Energy required to move Q coulomb charge through V volt as = Q.V = 50 × 3 = 150 J.
Correct Option: B
Energy required to move Q coulomb charge through V volt as = Q.V = 50 × 3 = 150 J.
- Calculate the value of current i—
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Current in branch BA, say i1
KCL at node B. i + i1 = 2
i1 = 1 amp.
Current in branch AC, say i2
KCL at node A:
6 = i1 + i2
6 = 1 + i2
i2 = 6 – 1 = 5 amp
Current in branch CD, say i3
KCL at node BC
4 + i2 = i3
4 + 5 = i3
or i3 = 9 amp current branch DE i.e. i
i + 8 = 9
i = 1ACorrect Option: A
Current in branch BA, say i1
KCL at node B. i + i1 = 2
i1 = 1 amp.
Current in branch AC, say i2
KCL at node A:
6 = i1 + i2
6 = 1 + i2
i2 = 6 – 1 = 5 amp
Current in branch CD, say i3
KCL at node BC
4 + i2 = i3
4 + 5 = i3
or i3 = 9 amp current branch DE i.e. i
i + 8 = 9
i = 1A
