Electric circuits miscellaneous
- In the figure below, the voltage across the 18 ohm resistor is 90 volts. What is the total voltage across the combined circuit?
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Since voltage across 18 ohm resistance is 90 volts, hence current through 18 ohm resistance
= 90 = 5 A 18
Equivalent resistance of 3 ohm and 6 ohm bank= 3 × 6 = 2 ohms 3 + 6
Since this 2 ohms resistance is in series with 18 ohm resistance, therefore Total resistance = 18 + 2 = 20 ohms
This 20 ohm resistance is in parallel with 5 ohm resistance∴ Equivalent resistance = 5 × 20 = 4 ohms 5 + 20
Hence total resistance of the circuit = 1 + 4 = 5 ohms. Current through this branch is 5 A
∴ Voltage across = 5 × 20 = 100 V = voltage across bc.
Hence, current through 5-ohm resistance= 100 = 20 A 5
∴ Total current through ab = 20 + 5 = 25 A. Since total resistance of the circuit is 5 ohm, therefore voltage, E = 25 × 5 = 125 V.Correct Option: A
Since voltage across 18 ohm resistance is 90 volts, hence current through 18 ohm resistance
= 90 = 5 A 18
Equivalent resistance of 3 ohm and 6 ohm bank= 3 × 6 = 2 ohms 3 + 6
Since this 2 ohms resistance is in series with 18 ohm resistance, therefore Total resistance = 18 + 2 = 20 ohms
This 20 ohm resistance is in parallel with 5 ohm resistance∴ Equivalent resistance = 5 × 20 = 4 ohms 5 + 20
Hence total resistance of the circuit = 1 + 4 = 5 ohms. Current through this branch is 5 A
∴ Voltage across = 5 × 20 = 100 V = voltage across bc.
Hence, current through 5-ohm resistance= 100 = 20 A 5
∴ Total current through ab = 20 + 5 = 25 A. Since total resistance of the circuit is 5 ohm, therefore voltage, E = 25 × 5 = 125 V.
- A certain network consists of two ideal voltage sources and a large number of ideal resistors. The power consumed in one of the resistor is 4 W when either of the two sources is active and the other is replaced by a short circuit. The power consumed by the same resistor when both the sources are simultaneously active would be
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The power consumed by the same resistor will be either 4 – 4 = 0 W; or 4 + 4 = 8 W.
Correct Option: C
The power consumed by the same resistor will be either 4 – 4 = 0 W; or 4 + 4 = 8 W.
- The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three circuit elements in watts is _____.
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NA
Correct Option: A
NA
- An ideal capacitor is charged to a voltage V0 and connected at t = 0 across an ideal inductor L. (The circuit
now consists of a capacitor and inductor alone). If we let ω0 = 1 √LC
the voltage across the capacitor at time t > 0 is given by
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Representation of charged capacitor to voltage V0 in S-domain. When this capacitor is connected to inductor at t > 0, the circuit becomes as,
I L (0+) = 0
and, VL (s) = I(s) · LsV(s) = V0 s s2 + 1 LC
Taking reverse Laplace, we get
v(t) = V0 cos ω0 t,ω0 = 1 √LC
Voltage across capacitor will discharge through inductor till voltage across capacitor is zero. When inductors becomes fully charged, it starts charging capacitor till current flow through the circuit is zero.
In this fashion, electrostatic energy gets exchanged into magnetic energy and vice-versa.
Correct Option: B
Representation of charged capacitor to voltage V0 in S-domain. When this capacitor is connected to inductor at t > 0, the circuit becomes as,
I L (0+) = 0
and, VL (s) = I(s) · LsV(s) = V0 s s2 + 1 LC
Taking reverse Laplace, we get
v(t) = V0 cos ω0 t,ω0 = 1 √LC
Voltage across capacitor will discharge through inductor till voltage across capacitor is zero. When inductors becomes fully charged, it starts charging capacitor till current flow through the circuit is zero.
In this fashion, electrostatic energy gets exchanged into magnetic energy and vice-versa.
- An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 H z, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater is removed from the supply and now connected to a 400 V peak to peak square wave source of 150 Hz. The power in kW dissipated by the heater will be
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Reading of Energy Meter = 2.3 Units (kWH)
2.3 × 103 = V2 × t = 230 × 230 × 1 R R
⇒ R = 23 Ω∴ Power dissipated in kW = V2 = 200 × 200 R R = 40 × 103 = 1.739 kW. 23
Correct Option: B
Reading of Energy Meter = 2.3 Units (kWH)
2.3 × 103 = V2 × t = 230 × 230 × 1 R R
⇒ R = 23 Ω∴ Power dissipated in kW = V2 = 200 × 200 R R = 40 × 103 = 1.739 kW. 23
