A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is
n 1020, Q = ne = e 1020 = 16.02 C.
Charge on sphere will be positive.
A lightning bolt carrying 15,000 A lasts for 100 μs. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is
ΔQ = i x Δ t = 15000 x 100μ = 1.5 C
If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is
The energy required to move 120 coulomb through 3 V is
W = Qv = 360 J
i = ?
In the circuit of fig P1.1.6 a charge of 600 C is delivered to the 100 V source in a 1 minute. The value of v1 must be
In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise.
Applying KVL we get
v1 + 60 - 100 = 10 x 20 or v1 = 240 V
In the circuit of the fig the value of the voltage source E is
Going from 10 V to 0 V
10+5+E+1 = 0
Consider the circuit graph shown in fig. Each branch of circuit graph represent a circuit element. The value of voltage v1 is
100 = 65 + V2 ⇒ v2 = 35 V
v3 - 30 = v2 ⇒ v3 = 65 V
105 + v4 - v3 - 65 = 0 ⇒ v4 = 25 V
v4 + 15 - 55 + v1 = 0 ⇒ v1 = 15 V
What quantity of charge must be delivered by a battery with a Potential Difference of 110 V to do 660J of Work?
R1 = ?
Voltage across 60 Ω resistor = 30 V Current = 30/60 =0.5 A
Voltage across R1 is =70 - 20 = 50 V
R1 = 50/0.5 =100Ω
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
v1 = ?
If we go from + side of 1 kΩ through 7 V, 6 V and
5 V, we get v1 = 7 + 6 - 5 = 8V
The voltage vo in fig is always equal to
It is not possible to determine the voltage across 1 A source.
Req = ?
We infer from the given diagram that the same pattern is followed after every 10ohm resistor. The infinite pattern in parallel as a whole is considered as to be Rx.
Thus, Rx = R + (R||Rx)
Solving for Rx, we get Rx = 1.62R, where R=10ohm.
So we have Rx = (1.62ohm)*(10ohm), which gives Rx=16.2ohm
Therefore, Req = 5 + (10||16.2)
=> 5 + [(10*16.2)/(10+16.2)] => Req = 5 + (162/26.2) which gives Req=11.18 ohm.
The quantity of a charge that will be transferred by a current flow of 10 A over 1 hour period is_________ ?
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
12C =2m x 10 C ⇒ 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
At t = 4 ms, vc = 0.05 V
It will be parabolic path. at t = 0 t-axis will be tangent.