Question

When a 22 ohms resistance is connected across its terminals a potential difference 2.2V is produced. Find the internal resistance and lost volt of the cell... Please solve this sum..This sum is from Current Electricity chapter...Class 10..Please I will mark you brainliest if you solve this sum...​

Answer 1

$\huge\red{ANSWER:-}$

Final Answer : 1.95V

Underst\nding :

1) Potential Difference across terminal of a cell,

V = E - I r

where

E = EMF of cell =2V

r = internal resistance =0.1ohm

i = current in circuit.

R = External Resistor

Steps :

1) Cell and External resistor are connected directly,

=> R(eq) = r + R

=0.1 + 3.9 = 4.0 oh m

2) Current in circuit,

I = E / R(eq)

= 2/4.0 = 1/2 =0.5 A

3) Potential Difference across terminal of a cell,

V = E - Ir

= 2 - 0.5 *0.1

= 1.95 V.

See Circuit Diagram in pic

Answer 2

Explanation:

P.D.=2.2volts , this is the EMF of the cell.

Now when the circuit is closed and a current is flowing through the circuit then the P.D.=1.8V

Thus the total resistance of the circuit is R+r=5+rohms

So, the current in the circuit is i=(

5+r

2.2

hence the PD across the internal resistance.

Given that,

5+r

2.2

)=2.2−1.8 as the drop in the potential across the battery is due the P.D. across the internal resistance.

So,

r(

5+r

2.2

)=0.4

1.8r=2

r=1.11ohms

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