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A solid copper sphere, 10 cm in diameter is deprived of 10^{20} electrons by a charging scheme. The charge on the sphere is
n = 10^{20} , Q = ne = 10^{20} 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
W = Qv = 360J
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 v_{1} 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 v_{1} is
For the circuit shown in fig P.1.1.9 the value of voltage v_{o} is
Voltage is constant because of 15 V source.
Voltage across 60 Ω resistor = 30 V
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