Q) 2 resistors of 5 KΩ and 7KΩ are connected in series across a 15V battery, then current flowing - Science - Force and Laws of Motion

Question

Q) 2 resistors of 5   K Ω   a n d   7 K Ω
are connected in series across a 15V battery, then current flowing
through each of resistance should be
(A) 2 mA
(B) 3 6 mA
(C) 5 mA
(D) 1 25 mA

Shruti Tyagi · Accepted Answer

Dear student,

Equivalent resistence when connected in series Re=R1+R2Re=5+7KΩRe=12KΩAs we know current in series remains same soBy ohm's lawV=IR⇒15=I×12⇒I=1512⇒I=54⇒I=1
25 mAhence option D is correct

Regards

Q). 2 resistors of 5 K Ω a n d 7 K Ω are connected in series across a 15V battery, then current flowing through each of resistance should be (A) 2 mA (B) 3.6 mA (C) 5 mA (D) 1.25 mA

Q). 2 resistors of $5 K Ω a n d 7 K Ω$ are connected in series across a 15V battery, then current flowing through each of resistance should be
(A) 2 mA
(B) 3.6 mA
(C) 5 mA
(D) 1.25 mA