A 9 volt battery is arranged in series combination with resistors of 2Ω, 3Ω, 4Ω, 5Ω and 12Ω, what would be the current flowing across the 2Ω resistor?
Answers
Answered by
2
V=RI
9=2×I
I=9/2
I=4.5 amp.
so, the current flowing through 2 ohm resistor is 4.5 ampere.
9=2×I
I=9/2
I=4.5 amp.
so, the current flowing through 2 ohm resistor is 4.5 ampere.
Answered by
4
Hello There !
__________________________________________________________
Here,
Potential(V) = 9V
Resistance(R) = 2Ω
Current (C) = ?
_________________________________________________________
As We Know,
I = V/R
I = 9/2
I = 4.5 Ampere
________________________________________________________
__________________________________________________________
Here,
Potential(V) = 9V
Resistance(R) = 2Ω
Current (C) = ?
_________________________________________________________
As We Know,
I = V/R
I = 9/2
I = 4.5 Ampere
________________________________________________________
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