Cells In Series And In Parallel

Given:
Total number of cells = 10,
EMF of each cell = $2%5C%2CV$ ,
Internal resistance of each cell = $1%5C%2C%5COmega$ ,
External resistance, $R%20%3D%2010%5C%2C%5COmega$ ,
Number of wrongly connected cells = 2.

Step-by-step Explanation:

Step 1: Determine the effective EMF of the battery combination.

In series, correctly connected cells add their EMFs, whereas wrongly connected cells subtract their EMFs.

Correctly connected cells = $10%20-%202%20%3D%208$

Net EMF of combination ( $E_%7Bnet%7D$ ): $E_%7Bnet%7D%20%3D%20(8%20%5Ctimes%202%5C%2CV)%20-%20(2%20%5Ctimes%202%5C%2CV)%20%3D%2016%5C%2CV%20-%204%5C%2CV%20%3D%2012%5C%2CV$

Step 2: Calculate total internal resistance:

All cells are in series, thus internal resistances add up directly. Total internal resistance ( $r_%7Btotal%7D$ ): $r_%7Btotal%7D%20%3D%2010%20%5Ctimes%201%5C%2C%5COmega%20%3D%2010%5C%2C%5COmega$

Step 3: Find the total resistance of the circuit:

Total resistance ( $R_%7Btotal%7D$ ) = Internal resistance + External resistance: $R_%7Btotal%7D%20%3D%20r_%7Btotal%7D%20%2B%20R%20%3D%2010%5C%2C%5COmega%20%2B%2010%5C%2C%5COmega%20%3D%2020%5C%2C%5COmega$

Step 4: Calculate the current ( $I$ ) through the resistor using Ohm’s law:

$I%20%3D%20%5Cfrac%7BE_%7Bnet%7D%7D%7BR_%7Btotal%7D%7D%20%3D%20%5Cfrac%7B12%5C%2CV%7D%7B20%5C%2C%5COmega%7D%20%3D%200.6%5C%2CA$

Final Conclusion:
The current through the resistor is 0.6 A.

We are given a 2 V battery with an internal resistance of 0.1

%5COmega

and an external resistance of 3.9

%5COmega

. The current

I

in the circuit can be calculated using Ohm's law:

I%20%3D%20%5Cfrac%7BV%7D%7BR_%7B%5Ctext%7Btotal%7D%7D%7D

where

V%20%3D%202%20%5C%2C%20%5Ctext%7BV%7D

is the battery voltage and

R_%7B%5Ctext%7Btotal%7D%7D%20%3D%20R_%7B%5Ctext%7Binternal%7D%7D%20%2B%20R_%7B%5Ctext%7Bexternal%7D%7D%20%3D%200.1%20%5C%2C%20%5COmega%20%2B%203.9%20%5C%2C%20%5COmega%20%3D%204.0%20%5C%2C%20%5COmega

. Thus, the current is:

I%20%3D%20%5Cfrac%7B2%7D%7B4.0%7D%20%3D%200.5%20%5C%2C%20%5Ctext%7BA%7D

Now, the voltage drop across the internal resistance is:

V_%7B%5Ctext%7Binternal%7D%7D%20%3D%20I%20%5Ctimes%20R_%7B%5Ctext%7Binternal%7D%7D%20%3D%200.5%20%5Ctimes%200.1%20%3D%200.05%20%5C%2C%20%5Ctext%7BV%7D

The terminal voltage across the cell is:

V_%7B%5Ctext%7Bterminal%7D%7D%20%3D%20V%20-%20V_%7B%5Ctext%7Binternal%7D%7D%20%3D%202.0%20%5C%2C%20%5Ctext%7BV%7D%20-%200.05%20%5C%2C%20%5Ctext%7BV%7D%20%3D%201.95%20%5C%2C%20%5Ctext%7BV%7D

Thus, the correct answer is

1.50%20%5C%2C%20%5Ctext%7BV%7D

About Cells In Series And In Parallel - KCET

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