Question

. When a 10 Ω resistor is connected across the terminals of a 4 V battery, the number of coulombs passing through the resistor per second is: *

Answer 1

Explanation:

Given-

R=10 ohm

p. d=4v

t=1 sec

find -

Q=?

Solution:-

v=IR

4=I×10

4/10=I

I=0.4A

I =Q/t

0.4=Q/1

Q=0.4 C

Ans- 0.4 C in second

Answer 2

PHYSICS

Given :-

R = 10 Ω

V = 4 v

T = 1 sec

According to Ohm's Law,

We know that :-

V = IR

Substitute the given values in the formula above to get the value of current.

4 = I * 10

I = 4/10

I = 0.4 A

Now, we know that

Q = IT

Substitute the values of current calculated and time to get the value of no. of Coulombs passing through the resistor in 1 second.

Q = 0.4 * 1

Q = 0.4 Coulombs

⇒ The number of Coulombs passing through the resistor per second is 0.4 Coulombs.