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A resistance of 9 ohm is connected to the terminals of a cell .A voltmeter connected across the cell reads 1.8V . When a resistance of 10 ohm is connected in series with 9ohm , the voltmeter reading changes to 1.9 V. Calculate the emf of the cell and it's internal resistance ?
Most Upvoted Answer
A resistance of 9 ohm is connected to the terminals of a cell .A voltm...
Solution:

Given:
Resistance of the circuit, R1 = 9 ohm
Resistance of the circuit with an additional resistor, R2 = 10 ohm
Voltage across the cell, V1 = 1.8V
Voltage across the cell with an additional resistor, V2 = 1.9V

To Find:
EMF of the cell, E
Internal resistance of the cell, r

Formula:
EMF of the cell, E = V1 + Ir
Voltage across the resistor, V = IR
Internal resistance of the cell, r = (E - V1) / I

Calculation:
According to the formula,
E = V1 + Ir
E = 1.8V + 9ohm*I --- Equation 1

When an additional resistor of 10 ohm is connected in series with the 9 ohm resistor, the voltage across the cell becomes V2 = 1.9V.

Now, the total resistance of the circuit, R = R1 + R2
R = 9ohm + 10ohm
R = 19ohm

The current passing through the circuit, I = V2 / R
I = 1.9V / 19ohm
I = 0.1A

From Equation 1,
E = 1.8V + 9ohm*I
E = 1.8V + 9ohm*0.1A
E = 1.8V + 0.9V
E = 2.7V

Internal resistance of the cell, r = (E - V1) / I
r = (2.7V - 1.8V) / 0.1A
r = 9ohm

Answer:
EMF of the cell, E = 2.7V
Internal resistance of the cell, r = 9ohm
Community Answer
A resistance of 9 ohm is connected to the terminals of a cell .A voltm...
£=I(R+r)=1.8v+0.2r. I=v/r=1.8v/9ohm=0.2A
£=I(R+r)=1.9v+0.1r. I=v/r=1.9v/9ohm=0.1A
1.8v+0.2r=1.9v+0.1r
r=1ohm
£=1.8v+0.2(1ohm)=2ohm
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A resistance of 9 ohm is connected to the terminals of a cell .A voltmeter connected across the cell reads 1.8V . When a resistance of 10 ohm is connected in series with 9ohm , the voltmeter reading changes to 1.9 V. Calculate the emf of the cell and it's internal resistance ?
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