Question 26:
(a) Define electric current. What is the SI unit of electric current.
(b) One coulomb of charge flows through any cross-section of a conductor in 1 second. What is the current flowing through the conductor ?
(c) Which instrument is used to measure electric current ? How should it be connected in a circuit ?
(d) What is the conventional direction of the flow of electric current ? How does it differ from the direction of flow of electrons ?
(e) A flash of lightning carries 10 C of charge which flows for 0.01 s. What is the current ? If the voltage is 10 MV, what is the energy ?
Lakhmir Singh Physics Class 10
Answers
Answer:
a. rmelectric current is defined as the rate of flow of electric charge...
s.i units....AMPERE (A)
b.One coulomb is nearly equal to 6 x 1018 electrons. SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second. This means if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.
c.ampere meter...or ammeter
d.The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.
e.Q=10 c,t=0.01 s
I=Q/T=10/0.01=1000A
p.d=W/Q=
W=p.d×Q
=10x10^6x10
=100MJ
Answer:
a. rmelectric current is defined as the rate of flow of electric charge...
s.i units....AMPERE (A)
b.One coulomb is nearly equal to 6 x 1018 electrons. SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second. This means if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.
c.ampere meter...or ammeter
d.The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.
e.Q=10 c,t=0.01 s
I=Q/T=10/0.01=1000A
p.d=W/Q=
W=p.d×Q
=10x10^6x10
=100MJ