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 = 1020 e (e = 1.6 x 10-19 )
= 16.02 C
The charge on the sphere is 16.02 coulomb. It 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
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 3V is
W = Qv = 360J
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
In the circuit of the fig P1.1.7, the value of the voltage source E is
Going from 10 V to 0 V
10 + 5 + E + 1 + 0 = 0 or E =-16 V
Consider the circuit graph shown in fig. P1.1.8. Each branch of circuit graph represent a circuit element. The value of voltage v1 is
For the circuit shown in fig P.1.1.9 the value of voltage vo is
Voltage is constant because of 15 V source.
R1 = ?
Voltage across 60 Ω resistor = 30 V