Question

A battery has an e.m.f. of 12.8V and supplies a current of 3.2A. What is the
resistance of the circuit? How many coulombs leave the battery in 5 minute?​

Answer 1

Given :-

Potential difference (V) = 12.8 v

Current (C) = 3.2 A

Time taken (T) = 5 min

To Find :-

Resistance (R)

Coulombs leave the battery in 5 minute?

Solution :-

Firstly let's convert the minutes into second

5 min = 5 × 60 = 300 sec

Now,

Let's find resistance by Ohm law

$\huge \sf \: V = IR$

Here,

V = Potential difference

I = Current

R = Resistance

12.8 = 3.2 × R

R = 12.8/3.2

R = 4 ohm

Hence the resistance of battery is 4 ohm.

Now,

Let's find charge

$\huge \sf \: I = Q \times T$

3.2 = Q × 300

3.2 = 300Q

Q = 300× 3.2

Q = 960 C

Hence the battery will leave 960 Columbous in 5 min.

Answer 2

Given    

emf = 12.8 V                    

I = 3.2 A              

t = 5 minutes = 5 * 60 = 300 seconds

Required      

 Resistance of the circuit, R =?                    

 Charge passed = Q =?

By Ohm’s law   V = IR .

Put the values

12.8=3.2R⇒R=12.8/3.2=4Ω

For finding the amount of charge Q, we use the definition of current.

I = Q/t

Put values and simplify,

3.2=Q/300⇒Q=3.2*300=960C

#SPJ3