A cell gives a current of 0.15A through a resistance of 8\(\Omega\) and 0.3A. When the resistance is changed to 3\(\Omega\) the internal resistance of the cell is
0.05\(\Omega\)
1.00\(\Omega\)
1.50\(\Omega\)
2.00\(\Omega\)
2.50\(\Omega\)
Explanation
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Make E the subject of formula from the equation I=E/R+r
You will have E=I(R+r)
However we are give two current and Resistor of which we are ask to find the internal resistance.....
Sine E=E
We have I(R+r)=I(R+r)
The values for each are
I=0.15A
R=8ohms............. equation 1
I=0.3A
R=3ohms............. equation 2
So we can deduce that
0.15(8+r)=0.3(3+r)
1.2+0.15r=0.9+0.3r
1.2-0.9=0.3r-0.15r
0.3=0.15r
r=0.3/0.15
r=2ohms
The answer is 1.5 ohms.
I= 0.15A
R= 8 ohms
I2= 0.3A
R2= 3 ohms
E= I( R + r)
E= 0.15(8+r)
E= 0.3(3+r)
Then, we have:
E = 1.2+ 0. 15r (eqn .... 1)
E= 0.9 + 0.3r (eqn ....2)
Collecting like terms
1.2+ 0.15r = 0.9+ 0.3r
1.2 - 0.9 = 0.15 - 0.3r
0.3 = 0.2r
r = 1.5 ohms
