Question

0.1. 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 ,I = 3.2 A
• Time taken ,t = 5 min = 5×60 = 300 s

To Find:-

• Resistance ,R
• Charge ,Q

Solution:-

Firstly we calculate the Resistance of the circuit.

Using Ohm's Law

•❒ V = IR

Where ,

• V denotes Voltage or Potential Difference

• I denotes current

• R denotes Resistance

Substitute the value we get

→ 12.8 = 3.2 × R

→ R = 12.8/3.2

→ R = 4 Ω

Therefore,the Resistance of the circuit is 4 ohms.

Now, We know that the relationship between Current , Charge & time .

❒ I = Q/t

Substitute the value we get

→ 3.2 = Q/300

→ Q = 3.2×300

→ Q = 960 C

Therefore,the charge leave the battery in 5 min is 960 Coulombs.

Answer 2

Answer:

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.