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
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