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For the circuit given below.
What is the value of V_{2}?
From Ohm’s law,
V_{2}= 3i and V_{1}=2i
Appling the KVL, 20 + V_{1} – V_{2}=0 ———–(i)
Put the value on above equation and get i = 4A
Hence, V_{2} = 3 x 4 = 12 V
For the circuit given below.
What is the value of equivalent resistance?
The circuit given below.The Thevenin resistance across the diode in the circuit is?
The Thevenin resistance across the diode in the circuit is?
Diode is nonlinear elements is removed.
After shorting the voltage source the redrawn circuit is
Thevenin resistance
= 11/3 ohm
Consider the following circuit.
What is the value of V_{24} / V_{13} of given circuit?
In the circuit given, a charge of 600 C is delivered to the 100 V source in a 1 minute. The value of v_{1} must be:
In the circuit of the fig the value of the voltage source E is:
Going from 10 V to 0 V:
Consider the circuit graph shown in fig. Each branch of circuit graph represent a circuit element. The value of voltage v_{1} is:
What quantity of charge must be delivered by a battery with a Potential Difference of 110 V to do 660J of Work?
► V = W / Q
► Q = W / V = 660 / 110 = 6 C
Twelve 6 Ω resistor are used as edge to form a cube.The resistance between two diagonally opposite corner of the cube is
The current i will be distributed in the cube branches symmetrically
If we go from +ve side of 1 kΩ through 7 V, 6 V and 5 V
We get, v_{1} = 7 + 6  5 = 8 V
The voltage v_{o} in fig is always equal to:
From above figure
V – 4 – 5 = 0
or
V = 9 V
Hence alternative (D) is the correct choice.
In a AC circuit, resistive and total impedance are 10 and 20 ohms respectively. What is the phase angle difference between the voltage and current?
The quantity of a charge that will be transferred by a current flow of 10 A over 1 hour period is _________ ?
We know that,
I=Q / t or Q = I x t = 10 x 1 hours
{since unit of current is Cs^{1}, therefore time should be in seconds}
∴ Q = 10 x 60 x 60
= 36000 C
= 3.6 x 10^{4 }C
A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec interval. The value of capacitance is:
12 * C = 2 m x 10
C = 1.67 mF
Hence, Capacitance is equal to 1.67 mF.
The energy required to charge a 10 μF capacitor to 100 V is:
The current in a 100 μF capacitor is shown in fig. If capacitor is initially uncharged, then the waveform for the voltage across it is:
This 0.2 V increases linearly from 0 to 0.2 V. Then current is zero. So capacitor hold this voltage.
The voltage across a 100 μF capacitor is shown in fig. The waveform for the current in the capacitor is:
= 600 mA
For 1 ms < t < 2 ms,
The waveform for the current in a 200 μF capacitor is shown in fig. The waveform for the capacitor voltage is:
= 3125 t^{2}
At t = 4 ms, v_{c} = 0.05 V
It will be parabolic path at t = 0
taxis will be tangent.
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