Question

b.) Twenty cells of internal resistance 0.5 ohm and emf 1.5V each are used to send a current

through an external resistance of 500 ohm. How would you arrange them to get the maximum

current.​

Answer 1

Question

Twenty cells of internal resistance 0.5 ohm and emf 1.5V each are used to send a current through an external resistance of 500 ohm. How would you arrange them to get the maximum current.?

Solution

Given that, number of cells = 20, Resistance = 0.5 ohm, emf = 1.5 and external resistance = 500 ohm

We have to find that how would we arrange them to get the maximum current.

As, the value of external resistance is more than that of internal resistance. So, to get the maximum current, we will arrange them in series.

We know that-

Current = (Total emf)/(Total resistance)

Total emf = nε

Here, n is the number of cells and ε is the emf

So, total emf = 20 × 1.5 = 30V

Total resistance = External resistance + (number of cells × Internal resistance)

Here, External resistance is denoted by R and internal resistance is denoted by r.

Total resistance = R + nr

= 500 + (20 × 0.5)

= 500 + 10

= 510 ohm

Therefore,

Current (I) = (Total emf)/(Total resistance)

Substitute the known values

I = 30/510

I = 0.059 A

Therefore, the maximum current is 0.059 A.

Answer 2

Question:

Twenty cells of internal resistance 0.5 ohm and emf 1.5V each are used to send a current through an external resistance of 500 ohm. How would you arrange them to get the maximum current?

Given:

1. Number of cells = 20
2. Resistance = 0.5 ohm & emf = 1.5
3. External resistance = 500 ohm

To Find:

• How would we arrange them to get the maximum current.

We know that,

$\pink{Current = \frac{Total \: emf}{Total \: resistance}}$

Total emf = nε

Here,

1. n = no. of cells
2. ε = emf.

So,

Total emf = 20 × 1.5

= 30V

→ Total resistance = External resistance (R) + (no. of cells × Internal resistance (r))

Here,

• External resistance = R
• Internal resistance = r

Total resistance = R + nr

→ 500 + (20 × 0.5)

→ 500 + 10

→ 510 ohm

Hence,

$\pink{Current(l) = \frac{Total \: emf}{Total \: resistance}}$

Substitute the known values are:-

1. I = $\frac{30}{510}$
2. I = 0.059A

Therefore,

The maximum value of current is 0.059A.